- Which of the following best describes what it means for the two-sided limit of f(x) as x approaches c to exist?
- The left-hand limit and the right-hand limit at c are both finite and equal to each other
- The function f is defined at x = c
- The function f has a value of zero at x = c
- The function f is increasing on both sides of c
Correct answer: The left-hand limit and the right-hand limit at c are both finite and equal to each other
A two-sided limit exists exactly when the left-hand and right-hand limits are both finite and equal to each other. The limit describes the single value f(x) approaches from both directions, so if the one-sided limits disagree the two-sided limit fails to exist. Whether f is actually defined at c, equals zero, or is increasing has no bearing on the existence of the limit.
- Evaluate lim x → 2 (x³ - 8)/(x - 2).
- 0
- 6
- 12
- The limit does not exist
Correct answer: 12
The limit equals 12. Factoring the difference of cubes gives ((x - 2)(x² + 2x + 4))/(x - 2), and canceling the common factor leaves x² + 2x + 4. Substituting x = 2 produces 4 + 4 + 4 = 12. The hole at x = 2 does not prevent the limit from existing.
- The function f given by f(x) = (x - 5)/(x² - 25) has a discontinuity at x = 5. What kind of discontinuity is it?
- A jump discontinuity
- An infinite discontinuity
- A removable discontinuity
- No discontinuity exists there
Correct answer: A removable discontinuity
The discontinuity at x = 5 is removable. Factoring the denominator as (x - 5)(x + 5) lets the common factor x - 5 cancel, leaving 1/(x + 5), whose limit as x → 5 is 1/10. Because the two-sided limit exists as a finite value, the gap is just a hole that could be filled, making it removable rather than a jump or infinite discontinuity.
- Suppose that for all x near 0, the inequality 4 - x² ≤ g(x) ≤ 4 + x² holds. What is lim x → 0 g(x)?
Correct answer: 4
The limit of g(x) is 4. Both bounding functions, 4 - x² and 4 + x², approach 4 as x → 0. By the squeeze theorem, a function trapped between two functions sharing a common limit must share that same limit, so g(x) is forced to 4.
- Evaluate lim x → 0 x cos(1/x).
- 0
- 1
- The limit equals cos(0)
- The limit does not exist
Correct answer: 0
The limit equals 0. Since cosine is always between -1 and 1, the product x cos(1/x) is trapped between -|x| and |x|. Both of those bounds approach 0 as x → 0, so the squeeze theorem forces the expression to 0 even though cos(1/x) oscillates wildly near zero.
- Evaluate lim x → ∞ (4x³ - 7x)/(2x³ + x² + 1).
Correct answer: 2
The limit equals 2. When the numerator and denominator have the same degree, the limit at infinity is the ratio of the leading coefficients, which here is 4/2. The lower-degree terms become negligible as x grows without bound, so the function levels off at the horizontal asymptote y = 2.
- Evaluate lim x → ∞ (5x + 2)/(x² - 3).
Correct answer: 0
The limit equals 0. The denominator is degree two while the numerator is degree one, so the denominator grows much faster than the numerator as x increases. When the degree of the denominator exceeds the degree of the numerator, the limit at infinity is 0, indicating a horizontal asymptote at y = 0.
- A continuous function f satisfies f(1) = -4 and f(3) = 6. Which conclusion about a root of f on the interval [1, 3] is guaranteed by the Intermediate Value Theorem?
- f has no roots on the interval
- f must equal zero at exactly one point on the interval
- There is at least one value c between 1 and 3 where f(c) = 0
- f equals zero only at the endpoints of the interval
Correct answer: There is at least one value c between 1 and 3 where f(c) = 0
The theorem guarantees at least one value c between 1 and 3 where f(c) = 0. Because f is continuous and changes sign from -4 to 6, it must pass through every intermediate value, including 0, somewhere between the endpoints. The theorem promises existence of such a c but does not specify how many roots there are or rule out additional ones.
- A polynomial function p is continuous everywhere with p(0) = -1 and p(2) = 7. Which value is the Intermediate Value Theorem guaranteed to produce somewhere on the interval [0, 2]?
Correct answer: 4
The value 4 is guaranteed because it lies between p(0) = -1 and p(2) = 7. The Intermediate Value Theorem ensures a continuous function attains every output value between its endpoint values. The numbers -3, 8, and 9 all fall outside the range from -1 to 7, so none of them is guaranteed.
- Why does the Intermediate Value Theorem require the function to be continuous on the closed interval before its conclusion can be applied?
- Continuity guarantees the function is also differentiable on the interval
- A discontinuity could let the function skip over intermediate values without attaining them
- Continuity ensures the function has a maximum value on the interval
- Continuity forces the function to be increasing across the interval
Correct answer: A discontinuity could let the function skip over intermediate values without attaining them
Continuity is required because a discontinuity could let the function jump over intermediate values without ever attaining them. The theorem relies on the unbroken nature of a continuous graph to ensure every value between the endpoints is hit; a break, such as a jump discontinuity, breaks that guarantee. Continuity by itself does not imply differentiability, the existence of a maximum, or monotonic behavior.
- Consider the piecewise function f where f(x) = 2x for x < 1 and f(x) = x + 4 for x ≥ 1. Does lim x → 1 f(x) exist?
- Yes, the limit equals 2
- Yes, the limit equals 5
- No, because the left-hand limit is 2 and the right-hand limit is 5
- No, because f is undefined at x = 1
Correct answer: No, because the left-hand limit is 2 and the right-hand limit is 5
The limit does not exist because the one-sided limits disagree. Approaching from the left, 2x tends to 2, while approaching from the right, x + 4 tends to 5. Since the left-hand limit of 2 does not equal the right-hand limit of 5, the two-sided limit fails to exist, which is the hallmark of a jump discontinuity.
- For the function f(x) = (x - 2)/(x² - 4), classify the discontinuities at x = 2 and at x = -2.
- Both are removable discontinuities
- Removable at x = 2 and infinite at x = -2
- Infinite at x = 2 and removable at x = -2
- Both are infinite discontinuities
Correct answer: Removable at x = 2 and infinite at x = -2
The discontinuity at x = 2 is removable while the one at x = -2 is infinite. Factoring the denominator as (x - 2)(x + 2) lets the numerator factor x - 2 cancel, leaving 1/(x + 2); this means x = 2 is just a hole with a finite limit of 1/4. At x = -2 nothing cancels, so the denominator goes to zero while the numerator does not, creating a vertical asymptote and an infinite discontinuity.
- What three conditions must hold for a function f to be continuous at x = a?
- f(a) is defined, lim x → a f(x) exists, and that limit equals f(a)
- f is increasing at a, f(a) is positive, and the limit exists
- f(a) = 0, f is differentiable at a, and the graph has no asymptote
- The left-hand limit exists, f(a) is defined, and f is bounded near a
Correct answer: f(a) is defined, lim x → a f(x) exists, and that limit equals f(a)
Continuity at a point requires that f(a) is defined, that lim x → a f(x) exists, and that the limit equals f(a). All three must hold together; if any one fails, the function is discontinuous there. Conditions like being increasing, positive, differentiable, or bounded are not part of the definition of continuity.
- Evaluate lim x → 0 (√(x + 9) - 3)/x.
- 0
- 1/3
- The limit does not exist
- 1/6
Correct answer: 1/6
The limit equals 1/6. Multiplying numerator and denominator by the conjugate √(x + 9) + 3 makes the numerator become x, which cancels the x in the denominator and leaves 1/(√(x + 9) + 3). Substituting x = 0 gives 1/(3 + 3) = 1/6.
- For what value of k is the function f, defined by f(x) = (x² - 9)/(x - 3) for x ≠ 3 and f(x) = k for x = 3, continuous at x = 3?
Correct answer: 6
The value k = 6 makes f continuous. Factoring the numerator as (x - 3)(x + 3) and canceling gives x + 3, whose limit as x → 3 is 6. To remove the discontinuity, k must equal that limiting value, so defining f(3) = 6 makes the limit equal the function value.
- It is known that lim x → 0 (sin x)/x = 1. Using this fact, evaluate lim x → 0 sin(5x)/x.
Correct answer: 5
The limit equals 5. Rewriting sin(5x)/x as 5 · sin(5x)/(5x) lets the inner expression sin(5x)/(5x) approach 1 as 5x → 0. Multiplying that limit of 1 by the factor of 5 gives 5.
- Evaluate lim x → 0 (3 - √(9 + x²))/x².
Correct answer: -1/6
The limit equals -1/6. Multiplying by the conjugate 3 + √(9 + x²) turns the numerator into 9 - (9 + x²), which simplifies to -x². Canceling x² leaves (-1)/(3 + √(9 + x²)), and substituting x = 0 gives (-1)/6.
- Evaluate lim x → 2⁺ (2 - x)/(|x - 2|).
- 1
- -1
- 0
- The limit does not exist
Correct answer: -1
The limit equals -1. For x slightly greater than 2, the quantity x - 2 is positive, so its absolute value is just x - 2. The numerator 2 - x equals -(x - 2), so the ratio simplifies to (-(x - 2))/(x - 2), which equals -1 for every x > 2.
- Evaluate lim x → ∞ (√(4x² + 1))/(x + 3).
Correct answer: 2
The limit equals 2. For large positive x, √(4x² + 1) behaves like √(4x²) = 2x, while the denominator behaves like x. Dividing the dominant terms (2x)/x gives 2, the horizontal asymptote as x grows without bound.
- A function f is continuous on the closed interval [2, 5] with f(2) = 8 and f(5) = 1. A student concludes there must be a value c in that interval where f(c) = 4. Is the reasoning valid, and why?
- No, because the Intermediate Value Theorem only applies when the function is increasing
- Yes, because 4 lies between f(2) and f(5) and f is continuous, so the Intermediate Value Theorem guarantees it
- No, because 4 is not the average of 8 and 1
- Yes, but only because f happens to be a polynomial
Correct answer: Yes, because 4 lies between f(2) and f(5) and f is continuous, so the Intermediate Value Theorem guarantees it
The reasoning is valid because 4 lies between the endpoint values 8 and 1 and f is continuous on the closed interval. The Intermediate Value Theorem guarantees that a continuous function attains every output value between its endpoint outputs, regardless of whether the function increases, is a polynomial, or relates to any average.
- For the function f(x) = 1/(x - 4), describe the behavior of lim x → 4 f(x) from the left and from the right.
- Both one-sided limits equal 0
- The left-hand limit is -∞ and the right-hand limit is +∞
- The left-hand limit is +∞ and the right-hand limit is -∞
- Both one-sided limits equal +∞
Correct answer: The left-hand limit is -∞ and the right-hand limit is +∞
The left-hand limit is -∞ and the right-hand limit is +∞. As x → 4⁻, x - 4 is a small negative number, so its reciprocal grows without bound in the negative direction. As x → 4⁺, x - 4 is a small positive number, so its reciprocal grows without bound in the positive direction, producing a vertical asymptote at x = 4.
- Suppose g is a function satisfying the inequality x² - 4x + 7 ≤ g(x) ≤ -x² + 4x - 1 for all real x. What is the value of g(2)?
- 7
- 2
- Cannot be determined from the information given
- 3
Correct answer: 3
The value of g(2) is 3. Substituting x = 2 into the lower bound x² - 4x + 7 gives 4 - 8 + 7 = 3, and into the upper bound -x² + 4x - 1 gives -4 + 8 - 1 = 3. Because the two bounds pinch together to the common value 3 at x = 2, the squeeze theorem forces g(2) to equal 3.
- Evaluate lim x → -∞ (7x² - 4x + 1)/(3 - 2x²).
Correct answer: -7/2
The limit equals -7/2. Because the numerator and denominator share the same degree of two, the limit at infinity equals the ratio of the leading coefficients, which is 7/(-2). The lower-degree terms vanish in the limit, so the function approaches the horizontal asymptote y = -7/2 from either infinite direction.
- Let f be continuous on the closed interval [-1, 4] with f(-1) = 5 and f(4) = -2. Which statement is guaranteed to be true by the Intermediate Value Theorem?
- f has a value of 6 somewhere on the interval
- f attains a maximum value of 5 on the interval
- f = 0 at some point between -1 and 4
- f is decreasing on the interval [-1, 4]
Correct answer: f = 0 at some point between -1 and 4
The guaranteed statement is that f = 0 at some point between -1 and 4. Since f is continuous and changes sign from 5 to -2, the value 0 lies between the endpoint outputs and must be attained by the Intermediate Value Theorem. The value 6 lies outside the range from -2 to 5, and the theorem says nothing about maximums or whether the function decreases.
- The derivative of a function f is defined as a limit. Geometrically, what does the value of f'(a) represent for the graph of f?
- The area enclosed between the graph and the x-axis up to x = a
- The vertical distance from the graph to the x-axis at x = a
- The slope of the line tangent to the graph at the point where x = a
- The average of all the function's outputs near x = a
Correct answer: The slope of the line tangent to the graph at the point where x = a
The value of f'(a) represents the slope of the line tangent to the graph at the point where x = a. The derivative is the limit of secant-line slopes as the second point slides toward a, and that limiting slope is the tangent-line slope. Area, vertical distance, and an average of outputs describe other quantities, not the instantaneous slope captured by the derivative.
- For the function f(x) = 1/x, the limit definition gives the difference quotient as (1/(x + h) - 1/x)/h. After combining the fractions in the numerator and simplifying, what is f'(x)?
Correct answer: -1/x²
The derivative is -1/x². Combining the numerator fractions over the common denominator x(x + h) gives -h in the top, and dividing by h cancels the h to leave (-1)/x(x + h). Letting h → 0 turns x + h into x, producing -1/x².
- A function f satisfies f(4) = 10, and its difference quotients near x = 4 approach the value -3 as the interval width shrinks to zero. What is f'(4)?
Correct answer: -3
The value of f'(4) is -3. The derivative at a point is precisely the limit that the difference quotients approach as the interval width goes to zero, and here that limit is stated to be -3. The output value 10 and the input 4 describe the location on the graph, not the rate of change there.
- Find the derivative of f(x) = 6³√(x).
- 2x^(-2/3)
- -2x^(-2/3)
- 6x^(-2/3)
- 2x^(2/3)
Correct answer: 2x^(-2/3)
The derivative is 2x^(-2/3). Rewriting ³√(x) as x^(1/3), the function is 6x^(1/3), and the power rule brings down 1/3 to give 6 · 1/3 = 2 and reduces the exponent to -2/3. The coefficient becomes 2 because 6 · 1/3 = 2.
- What is the derivative of f(x) = 8x?
Correct answer: 8
The derivative is 8. The function 8x is a linear function written as 8x¹, and the power rule gives 8 · 1 · x⁰ = 8 · 1 = 8. This matches the fact that a line of the form 8x has constant slope 8 everywhere.
- Find the second derivative of f(x) = x⁴ - 5x².
- 12x² - 10
- 4x³ - 10x
- 12x² - 10x
- 12x³ - 10
Correct answer: 12x² - 10
The second derivative is 12x² - 10. Differentiating once with the power rule gives 4x³ - 10x, and differentiating that result again gives 12x² - 10, since the derivative of 4x³ is 12x² and the derivative of -10x is -10. Applying the power rule twice is what produces the second derivative.
- Find the derivative of f(x) = (x² + 3)√x using the product rule.
- 2x√x
- 5/2x^(3/2) + 3/2x^(-1/2)
- 2x · x^(1/2)
- 2x^(3/2) - 3/2x^(-1/2)
Correct answer: 5/2x^(3/2) + 3/2x^(-1/2)
The derivative is 5/2x^(3/2) + 3/2x^(-1/2). The product rule gives the derivative of x² + 3, which is 2x, times x^(1/2), plus x² + 3 times the derivative of x^(1/2), which is 1/2x^(-1/2). The first term is 2x^(3/2) and the second contributes 1/2x^(3/2) + 3/2x^(-1/2), and combining the like x^(3/2) terms gives 5/2x^(3/2).
- Let f(x) = xtan x. Using the product rule together with the derivative of the tangent function, find f'(x).
- tan x + xsec² x
- sec² x
- xsec² x
- tan x - xsec² x
Correct answer: tan x + xsec² x
The derivative is tan x + xsec² x. The product rule gives the derivative of x, which is 1, times tan x, plus x times the derivative of tan x, which is sec² x. The first term is simply tan x and the second is xsec² x, and the two terms cannot be combined further.
- Evaluate the derivative of f(x) = x² eˣ at x = 0 using the product rule.
Correct answer: 0
The derivative at x = 0 is 0. The product rule gives the derivative of x², which is 2x, times eˣ, plus x² times the derivative of eˣ, which is eˣ, producing 2x eˣ + x² eˣ. Substituting x = 0 makes both terms contain a factor of x, so each term is zero and the sum is 0.
- Find the derivative of f(x) = (4x - 1)/(2x + 3) using the quotient rule.
- 14/((2x + 3)²)
- 2/((2x + 3)²)
- (-14)/((2x + 3)²)
- 14/(2x + 3)
Correct answer: 14/((2x + 3)²)
The derivative is 14/((2x + 3)²). The quotient rule gives the denominator 2x + 3 times the derivative of 4x - 1, which is 4, minus 4x - 1 times the derivative of 2x + 3, which is 2, all over (2x + 3)². The numerator is (8x + 12) - (8x - 2), which simplifies to 14.
- Find the derivative of f(x) = (cos x)/x using the quotient rule.
- -(sin x)/x²
- (-x sin x - cos x)/x²
- (x sin x + cos x)/x²
- (-x sin x + cos x)/x²
Correct answer: (-x sin x - cos x)/x²
The derivative is (-x sin x - cos x)/x². The quotient rule gives the denominator x times the derivative of cos x, which is -sin x, minus cos x times the derivative of x, which is 1, all over x². The numerator becomes -x sin x - cos x.
- A student rewrites f(x) = (6x³)/(2x) as 3x² before differentiating, while a classmate insists the quotient rule is required. Which approach is valid and what is the derivative?
- Only the quotient rule works; the derivative is 6x
- Both approaches give the same result; the derivative is 6x
- The rewrite is invalid because it changes the function; the derivative is 3x
- Both approaches give the same result; the derivative is 3x²
Correct answer: Both approaches give the same result; the derivative is 6x
Both approaches give the same result, and the derivative is 6x. Simplifying (6x³)/(2x) to 3x² is valid for all x ≠ 0, and the power rule then gives 6x directly. The quotient rule applied to the original form yields the identical answer after simplification, so rewriting first is simply the more efficient route.
- In the limit definition of the derivative at a fixed point a, written as lim h → 0 (f(a+h) - f(a))/h, what is changing as the limit is taken?
- The point a is sliding along the x-axis
- The width h of the interval is shrinking toward zero
- The output f(a) is growing without bound
- The function f is being replaced by a different function
Correct answer: The width h of the interval is shrinking toward zero
The quantity changing is the width h of the interval, which shrinks toward zero. As h → 0, the second input a+h slides toward the fixed point a, so the secant line through those two points approaches the tangent line. The point a stays fixed and the function f is unchanged; only the spacing h between the two inputs vanishes.
- Using the limit definition of the derivative, find f'(x) for f(x) = 4x² - 1.
Correct answer: 8x
The derivative is 8x. The difference quotient gives (4(x+h)² - 1 - (4x² - 1))/h, which expands to (8xh + 4h²)/h. Canceling h leaves 8x + 4h, and letting h → 0 yields 8x. The constant -1 contributes nothing because it cancels in the numerator.
- A table gives values of a function g, with g(2) = 7 and g(2.001) = 7.012. Which value best estimates g'(2) using a difference quotient?
Correct answer: 12
The best estimate of g'(2) is 12. Forming the difference quotient with the two given points uses the change in output, 7.012 - 7 = 0.012, divided by the change in input, 2.001 - 2 = 0.001. Dividing 0.012/0.001 gives 12, approximating the instantaneous rate of change with a very small interval width.
- Find the derivative of f(x) = 9/(√x).
- 9/2 x^(-1/2)
- -9 x^(-3/2)
- -9/2 x^(-3/2)
- -9/2 x^(-1/2)
Correct answer: -9/2 x^(-3/2)
The derivative is -9/2 x^(-3/2). Rewriting 9/(√x) as 9 x^(-1/2), the power rule brings down -1/2 to give 9 · (-1/2) = -9/2, and reduces the exponent by one to -3/2. The negative exponent reflects the function decreasing as x grows.
- Find the derivative of f(x) = 2x⁵ - 7x² + 6x.
- 10x⁴ - 14x + 6
- 10x⁴ - 7x + 6
- 2x⁴ - 14x + 6
- 10x⁴ - 14x
Correct answer: 10x⁴ - 14x + 6
The derivative is 10x⁴ - 14x + 6. Applying the power rule term by term, the derivative of 2x⁵ is 10x⁴, the derivative of -7x² is -14x, and the derivative of 6x is the constant 6. The sum and constant-multiple rules combine these term derivatives.
- The position of an object is modeled by s(x) = x⁴. Which expression gives the slope of the tangent line to the graph of s at any value x, using the power rule?
Correct answer: 4x³
The slope of the tangent line is 4x³. The derivative of a function gives the slope of its tangent line at each input, and the power rule on x⁴ brings down the exponent 4 and reduces the power to 3. Evaluating 4x³ at a particular x gives the tangent slope there.
- Let f(x) = x³ ln x. Using the product rule, find f'(x).
- 3x² ln x
- x² (3 ln x - 1)
- 3x² + 1/x
- x² (3 ln x + 1)
Correct answer: x² (3 ln x + 1)
The derivative is x² (3 ln x + 1). The product rule gives the derivative of x³, which is 3x², times ln x, plus x³ times the derivative of ln x, which is 1/x. That second term is x³/x = x², and factoring x² from both terms leaves 3 ln x + 1.
- Let f(x) = eˣ cos x. Using the product rule, find f'(x).
- -eˣ sin x
- eˣ (cos x + sin x)
- eˣ (cos x - sin x)
- eˣ cos x · (-sin x)
Correct answer: eˣ (cos x - sin x)
The derivative is eˣ (cos x - sin x). The product rule gives the derivative of eˣ, which is eˣ, times cos x, plus eˣ times the derivative of cos x, which is -sin x. Factoring out the common eˣ leaves cos x - sin x.
- Find the derivative of f(x) = (sin x)/x² using the quotient rule.
- (x² cos x - 2x sin x)/x⁴
- (cos x)/(2x)
- (x cos x - 2 sin x)/x³
- (cos x - 2 sin x)/x³
Correct answer: (x cos x - 2 sin x)/x³
The derivative is (x cos x - 2 sin x)/x³. The quotient rule gives the denominator x² times the derivative of sin x, which is cos x, minus sin x times the derivative of x², which is 2x, all over (x²)² = x⁴. The numerator x² cos x - 2x sin x and the denominator x⁴ share a factor of x, which cancels to give (x cos x - 2 sin x)/x³.
- Find the derivative of f(x) = x²/(x - 3) using the quotient rule, then evaluate it at x = 1.
Correct answer: -5/4
The derivative at x = 1 is -5/4. The quotient rule gives ((x-3)(2x) - x²(1))/((x-3)²) = (x² - 6x)/((x-3)²). At x = 1 the numerator is 1 - 6 = -5, and the denominator is (1 - 3)² = 4, giving -5/4.
- A student differentiates f(x) = x² cos x by multiplying the separate derivatives to get 2x · (-sin x) = -2x sin x. Why is this incorrect, and what is the correct derivative?
- The result is correct; the derivative is -2x sin x
- The derivative of a product is not the product of derivatives; the correct derivative is 2x cos x - x² sin x
- The student should have used the quotient rule; the correct derivative is 2x cos x
- The derivatives should be added, giving 2x - sin x
Correct answer: The derivative of a product is not the product of derivatives; the correct derivative is 2x cos x - x² sin x
The correct derivative is 2x cos x - x² sin x, and the error is treating the derivative of a product as the product of derivatives. The product rule requires the derivative of x², which is 2x, times cos x, plus x² times the derivative of cos x, which is -sin x. Combining these gives 2x cos x - x² sin x, not the single product the student computed.
- A function f is differentiable at x = a. Which statement must therefore be true about f at x = a?
- f has a horizontal tangent line at x = a
- f is continuous at x = a
- f has a relative extremum at x = a
- The second derivative of f exists at x = a
Correct answer: f is continuous at x = a
The statement that must be true is that f is continuous at x = a, since differentiability at a point guarantees continuity there. The limit defining the derivative can exist only if the graph is unbroken at that point. The other options are not forced: the tangent need not be horizontal, no extremum is required, and the existence of the first derivative does not guarantee the second.
- Find the derivative of f(x) = e^(3x²).
- e^(3x²)
- 6x e⁶ˣ
- 3x² e^(3x²)
- 6x e^(3x²)
Correct answer: 6x e^(3x²)
The derivative is 6x e^(3x²). By the chain rule, the derivative of an exponential e^u equals that same exponential multiplied by the derivative of the inner exponent. The exponent 3x² has derivative 6x, which becomes the leading factor while the exponential part is unchanged.
- A table gives f(3) = 4, f'(3) = 2, g(3) = 5, and g'(4) = 7. If h(x) = g(f(x)), what is h'(3)?
Correct answer: 14
The value of h'(3) is 14. The chain rule says h'(x) = g'(f(x)) · f'(x), so at x = 3 this is g'(f(3)) · f'(3). Since f(3) = 4, this becomes g'(4) · f'(3) = 7 · 2 = 14.
- Find the derivative of f(x) = ln(5x).
Correct answer: 1/x
The derivative is 1/x. By the chain rule, the derivative of ln of an inner function is the inner derivative divided by the inner function, giving 5/(5x). The 5 in the numerator and the 5 in the denominator cancel, leaving 1/x. Equivalently, ln(5x) = ln 5 + ln x, whose derivative is simply 1/x.
- Find the derivative of f(x) = tan(x³).
- sec²(x³)
- 3x² tan(x³)
- 3x² sec²(x³)
- sec²(3x²)
Correct answer: 3x² sec²(x³)
The derivative is 3x² sec²(x³). The chain rule gives the outer derivative of tangent, which is sec² evaluated at the inner function x³, multiplied by the inner derivative 3x². The argument x³ stays inside the sec² while 3x² appears as the front factor.
- Find the slope of the tangent line to the curve y² = 4x at the point (1, 2).
Correct answer: 1
The slope of the tangent line is 1. Differentiating y² = 4x implicitly gives 2y dy/dx = 4, so dy/dx = 4/(2y) = 2/y. Substituting y = 2 from the point produces 2/2 = 1.
- For the curve x³ + y³ = 6xy, find dy/dx using implicit differentiation.
- (2y - x²)/(y² - 2x)
- (x² - 2y)/(2x - y²)
- -x²/y²
- (6y - 3x²)/(3y²)
Correct answer: (2y - x²)/(y² - 2x)
The derivative dy/dx = (2y - x²)/(y² - 2x). Differentiating gives 3x² + 3y² dy/dx on the left, and on the right the product rule on 6xy gives 6y + 6x dy/dx. Collecting the dy/dx terms and dividing both numerator and denominator by 3 produces the stated quotient.
- For the curve sin(y) = x, find dy/dx using implicit differentiation.
- cos(y)
- 1/cos(y)
- -sin(y)
- 1/sin(y)
Correct answer: 1/cos(y)
The derivative dy/dx = 1/cos(y). Differentiating both sides with respect to x, the left side requires the chain rule and becomes cos(y) dy/dx, while the right side is 1. Solving cos(y) dy/dx = 1 isolates dy/dx = 1/cos(y).
- For the curve e^y + xy = 1, find dy/dx using implicit differentiation.
- (-y)/(e^y + x)
- y/(e^y + x)
- (-y)/x
- (1 - y)/(e^y)
Correct answer: (-y)/(e^y + x)
The derivative dy/dx = (-y)/(e^y + x). Differentiating e^y gives e^y dy/dx, and the product xy gives y + x dy/dx. Setting the sum equal to 0 and grouping the dy/dx terms yields dy/dx(e^y + x) = -y, so dy/dx = (-y)/(e^y + x).
- What is the derivative of f(x) = arctan x, the inverse tangent of x?
- 1/(√(1 - x²))
- 1/(1 + x²)
- -1/(1 + x²)
- 1/(1 - x²)
Correct answer: 1/(1 + x²)
The derivative of arctan x is 1/(1 + x²). This is the standard inverse-trigonometric derivative for the inverse tangent, and unlike the inverse sine it involves a sum rather than a square root. The form 1/(√(1 - x²)) belongs to arcsin x, which is a common point of confusion.
- Let f be a differentiable one-to-one function with inverse g. Given that f(0) = 1 and f'(0) = 8, find g'(1).
Correct answer: 1/8
The value of g'(1) is 1/8. The inverse function derivative rule states that the slope of the inverse at an output equals the reciprocal of the original function's slope at the matching input. Since f(0) = 1, the matching input is 0, so g'(1) = 1/(f'(0)) = 1/8.
- Find the derivative of f(x) = x arctan x.
- arctan x + x/(1 + x²)
- 1/(1 + x²)
- arctan x - x/(1 + x²)
- x/(1 + x²)
Correct answer: arctan x + x/(1 + x²)
The derivative is arctan x + x/(1 + x²). The product rule combines the derivative of x, which is 1, times arctan x, with x times the derivative of arctan x. Since the derivative of arctan x is 1/(1 + x²), the second term becomes x/(1 + x²), added to the first term arctan x.
- If g is the inverse of f and the point (3, 7) lies on the graph of f where the tangent line has slope 5, what is the slope of the tangent line to the graph of g at the point (7, 3)?
Correct answer: 1/5
The slope of the tangent to the graph of g at (7, 3) is 1/5. The graph of an inverse function reflects the original across the line y = x, which swaps the roles of input and output and inverts the slope. Because f has slope 5 at the corresponding point, the inverse g has the reciprocal slope of 1/5 at the reflected point.
- Find the derivative of f(x) = √(x² + 1).
- x/(√(x² + 1))
- 1/(√(x² + 1))
- (2x)/(√(x² + 1))
- √(2x)
Correct answer: x/(√(x² + 1))
The derivative is x/(√(x² + 1)). Writing the function as (x² + 1)^(1/2), the chain rule gives 1/2(x² + 1)^(-1/2) multiplied by the inner derivative 2x. The factor of 1/2 cancels with the 2 from 2x, leaving x/(√(x² + 1)).
- Find the derivative of f(x) = sin(4x).
- cos(4x)
- -4 cos(4x)
- 4 cos(4x)
- 4 sin(4x)
Correct answer: 4 cos(4x)
The derivative is 4 cos(4x). By the chain rule, the derivative of sine of an inner function is cosine of that same inner function times the inner derivative. The outer derivative gives cos(4x), and the inner function 4x has derivative 4, which multiplies in front to produce 4 cos(4x).
- Find the derivative of f(x) = (2x + 1)⁵.
- 5(2x + 1)⁴
- 10(2x + 1)⁴
- 2(2x + 1)⁴
- 10(2x + 1)⁵
Correct answer: 10(2x + 1)⁴
The derivative is 10(2x + 1)⁴. The chain rule brings down the exponent 5 and reduces the power by one, giving 5(2x + 1)⁴, then multiplies by the inner derivative of 2x + 1, which is 2. Multiplying 5 · 2 produces the coefficient 10.
- If h(x) = cos(x²), what is h'(x)?
- -sin(x²)
- -2x sin(x²)
- 2x sin(x²)
- -2x cos(x²)
Correct answer: -2x sin(x²)
The derivative is -2x sin(x²). The chain rule applied to cosine of an inner function yields negative sine of that inner function times the inner derivative. The outer step gives -sin(x²), and the inner function x² contributes a derivative of 2x, producing -2x sin(x²).
- Suppose u(2) = 3, u'(2) = 5, and v'(3) = 4. If k(x) = v(u(x)), what is k'(2)?
Correct answer: 20
The value of k'(2) is 20. The chain rule gives k'(x) = v'(u(x)) · u'(x), so at x = 2 this is v'(u(2)) · u'(2). Since u(2) = 3, this becomes v'(3) · u'(2) = 4 · 5 = 20.
- Find the slope of the tangent line to the circle x² + y² = 25 at the point (3, 4).
Correct answer: -3/4
The slope of the tangent line is -3/4. Differentiating x² + y² = 25 implicitly gives 2x + 2y dy/dx = 0, so dy/dx = -x/y. Substituting the point (3, 4) gives -3/4.
- For the curve xy = 8, find dy/dx using implicit differentiation.
Correct answer: -y/x
The derivative dy/dx = -y/x. Differentiating xy = 8 with the product rule gives y + x dy/dx = 0, since the derivative of the constant 8 is zero. Solving for dy/dx isolates it as -y/x.
- For the curve x² + xy + y² = 7, find dy/dx using implicit differentiation.
- -(2x + y)/(x + 2y)
- (2x + y)/(x + 2y)
- -(2x + y)/(2y - x)
- -x/y
Correct answer: -(2x + y)/(x + 2y)
The derivative dy/dx = -(2x + y)/(x + 2y). Differentiating term by term gives 2x, then the product rule on xy gives y + x dy/dx, and y² gives 2y dy/dx. Grouping the dy/dx terms gives dy/dx(x + 2y) = -(2x + y), so dy/dx = -(2x + y)/(x + 2y).
- What is the derivative of f(x) = arcsin x, the inverse sine of x?
- 1/(1 + x²)
- 1/(√(1 - x²))
- -1/(√(1 - x²))
- 1/(√(x² - 1))
Correct answer: 1/(√(1 - x²))
The derivative of arcsin x is 1/(√(1 - x²)). This is the standard inverse-trigonometric derivative for the inverse sine. The form 1/(1 + x²) belongs instead to the inverse tangent, which is a frequent source of confusion.
- Let f be a one-to-one differentiable function with inverse g, where f(2) = 6 and f'(2) = 3. What is g'(6)?
Correct answer: 1/3
The value of g'(6) is 1/3. The inverse function derivative rule says the slope of the inverse at an output equals the reciprocal of the original function's slope at the matching input. Since f(2) = 6, the matching input is 2, so g'(6) = 1/(f'(2)) = 1/3.
- Find the derivative of f(x) = e^(sin x).
- cos x · e^(sin x)
- e^(sin x)
- sin x · e^(sin x)
- -cos x · e^(sin x)
Correct answer: cos x · e^(sin x)
The derivative is cos x · e^(sin x). By the chain rule, the derivative of e^u equals e^u times the derivative of the exponent. The exponent sin x has derivative cos x, which becomes the front factor while the exponential e^(sin x) is unchanged.
- For the curve y = arctan(2x), find dy/dx.
- 1/(1 + 4x²)
- 2/(1 + 4x²)
- 2/(1 + 2x²)
- 1/(1 + 2x)
Correct answer: 2/(1 + 4x²)
The derivative is 2/(1 + 4x²). The inverse tangent derivative gives 1/(1 + (inner)²), where the inner function is 2x, so the denominator is 1 + (2x)² = 1 + 4x². The chain rule then multiplies by the inner derivative 2, placing a 2 in the numerator.
- In a related rates problem, what relationship is being analyzed?
- How the average value of a function compares to its endpoints
- How the rate of change of one quantity relates to the rate of change of another quantity that shares an equation with it
- How the area under a curve accumulates over time
- How a function behaves as the input grows without bound
Correct answer: How the rate of change of one quantity relates to the rate of change of another quantity that shares an equation with it
A related rates problem analyzes how the rate of change of one quantity relates to the rate of change of another quantity linked to it by an equation. By differentiating the relating equation with respect to time, each variable's time derivative appears, so a known rate determines an unknown rate. The other choices describe average value, accumulation, and end behavior, which are different ideas entirely.
- The radius of a circle is increasing at a rate of 3 centimeters per second. How fast is the area increasing at the moment the radius is 5 centimeters?
- 30π square centimeters per second
- 15π square centimeters per second
- 25π square centimeters per second
- 10π square centimeters per second
Correct answer: 30π square centimeters per second
The area is increasing at 30π square centimeters per second. Differentiating the area formula A = π r² with respect to time gives (dA)/dt = 2π r · dr/dt. Substituting r = 5 and dr/dt = 3 yields 2π · 5 · 3 = 30π.
- A 13-foot ladder leans against a vertical wall, and its base is pulled away from the wall at 2 feet per second. When the base is 5 feet from the wall, how fast is the top of the ladder sliding down?
- 5/6 of a foot per second downward
- 2 feet per second downward
- 5/12 of a foot per second downward
- 12/5 of a foot per second downward
Correct answer: 5/6 of a foot per second downward
The top slides down at 5/6 of a foot per second. With x² + y² = 169, when x = 5 the height y = 12. Differentiating gives 2x · dx/dt + 2y · dy/dt = 0, so dy/dt = -(x · dx/dt)/y = -(5 · 2)/12 = -5/6, the negative sign indicating downward motion.
- What does the linear approximation of a function f at x = a use to estimate nearby function values?
- The secant line connecting two distant points on the graph
- The horizontal asymptote of the function
- The average rate of change over a wide interval
- The tangent line to the graph of f at the point where x = a
Correct answer: The tangent line to the graph of f at the point where x = a
Linear approximation uses the tangent line to the graph of f at x = a to estimate nearby values. Because a differentiable curve closely follows its tangent line for inputs near the point of tangency, the line's outputs approximate the function's outputs. A secant line, an asymptote, or an average rate of change over a wide interval would not give the local tangent-based estimate.
- Use a tangent-line approximation of f(x) = √x at x = 25 to estimate √(26).
Correct answer: 5.1
The estimate is 5.1. The linear approximation is f(25) + f'(25) · Δ x, where f(25) = 5 and f'(x) = 1/(2√x), giving f'(25) = 1/10. With a change of 1 from 25 to 26, the estimate is 5 + 1/10 · 1 = 5.1.
- Evaluate lim x → 0 (sin x)/(3x) using L'Hopital's Rule.
Correct answer: 1/3
The limit equals 1/3. Direct substitution gives the indeterminate form 0/0, so L'Hopital's Rule allows differentiating numerator and denominator separately, producing (cos x)/3. Evaluating (cos x)/3 at x = 0 gives 1/3.
- Evaluate lim x → 0 (e²ˣ - 1)/x using L'Hopital's Rule.
Correct answer: 2
The limit equals 2. Substitution gives the indeterminate form 0/0, so L'Hopital's Rule applies by differentiating top and bottom to get (2e²ˣ)/1. Evaluating at x = 0 gives 2 · e⁰ = 2 · 1 = 2.
- Before applying L'Hopital's Rule to a limit, which condition must first be verified?
- The function must be increasing on the interval
- The numerator and denominator must be polynomials
- The numerator and denominator must be equal at the point
- The limit must produce an indeterminate form such as 0/0 or ∞/∞
Correct answer: The limit must produce an indeterminate form such as 0/0 or ∞/∞
The rule may be applied only after confirming the limit yields an indeterminate form such as 0/0 or ∞/∞. L'Hopital's Rule is justified solely for these forms, and using it on a determinate limit can give a wrong answer. The functions need not be polynomials or increasing, and equal values are not the requirement.
- A particle moves along a line with position s(t) = t³ - 6t² + 9t. What is the particle's velocity at time t = 2?
Correct answer: -3
The velocity at t = 2 is -3. Velocity is the derivative of position, so differentiating gives v(t) = 3t² - 12t + 9. Substituting t = 2 produces 3 · 4 - 12 · 2 + 9 = 12 - 24 + 9 = -3.
- A particle moving along a line has velocity v(t) = t² - 4t + 3. At which time is the particle momentarily at rest while changing direction from moving right to moving left?
Correct answer: t = 1
The particle changes from rightward to leftward motion at t = 1. Setting v(t) = 0 gives t = 1 and t = 3, the candidate times for being at rest. For t just below 1 the velocity is positive (moving right) and just above 1 it is negative (moving left), so the direction reverses from right to left at t = 1, whereas at t = 3 it reverses from left to right.
- The volume of a cube is increasing at a rate of 24 cubic centimeters per second. How fast is each edge length increasing at the instant the edge length is 2 centimeters?
- 6 centimeters per second
- 12 centimeters per second
- 2 centimeters per second
- 1/2 centimeter per second
Correct answer: 2 centimeters per second
Each edge is increasing at 2 centimeters per second. Differentiating the volume V = s³ gives (dV)/dt = 3s² · ds/dt. Substituting (dV)/dt = 24 and s = 2 gives 24 = 3 · 4 · ds/dt, so 24 = 12 · ds/dt, making ds/dt = 2.
- A spherical balloon is being inflated so that its volume increases at 100 cubic centimeters per second. Using the relationship (dV)/dt = 4π r² · dr/dt, how fast is the radius increasing when the radius is 5 centimeters?
- 1/π centimeters per second
- 1/(2π) centimeters per second
- 1/(5π) centimeters per second
- 4/π centimeters per second
Correct answer: 1/π centimeters per second
The radius is increasing at 1/π centimeters per second. Substituting (dV)/dt = 100 and r = 5 into (dV)/dt = 4π r² · dr/dt gives 100 = 4π · 25 · dr/dt = 100π · dr/dt. Dividing both sides by 100π leaves dr/dt = 1/π.
- Use a linear approximation to estimate (2.01)³, basing the approximation at x = 2 for f(x) = x³.
Correct answer: 8.12
The estimate is 8.12. With f(x) = x³, the value f(2) = 8 and the derivative f'(x) = 3x² gives f'(2) = 12. The linear approximation is 8 + 12 · 0.01 = 8 + 0.12 = 8.12.
- A student tries to evaluate lim x → 3 (x² - 9)/(x - 3) and immediately applies L'Hopital's Rule to get (2x)/1 = 6. Although the answer is correct, why was using the rule not strictly necessary here?
- The limit was actually ∞/∞, not 0/0
- The expression factors and simplifies to x + 3, removing the indeterminate form algebraically
- L'Hopital's Rule never applies to rational functions
- The limit does not exist, so any method fails
Correct answer: The expression factors and simplifies to x + 3, removing the indeterminate form algebraically
The rule was unnecessary because the numerator factors as (x - 3)(x + 3), so canceling the common factor x - 3 simplifies the expression to x + 3, whose limit at x = 3 is 6. The original form is genuinely 0/0, but algebraic simplification resolves it directly, making L'Hopital's Rule an optional shortcut rather than a requirement here.
- For a particle moving along a line, what does the sign of the acceleration compared with the sign of the velocity tell you?
- When acceleration and velocity have the same sign, the particle is speeding up; when they have opposite signs, it is slowing down
- When acceleration is positive, the particle is always moving to the right
- When velocity is zero, the particle must be at the origin
- When acceleration equals velocity, the particle stops permanently
Correct answer: When acceleration and velocity have the same sign, the particle is speeding up; when they have opposite signs, it is slowing down
A particle is speeding up when acceleration and velocity share the same sign and slowing down when their signs differ, because speed is the magnitude of velocity. Same-sign means the velocity is growing in magnitude, while opposite signs mean it is shrinking toward zero. Positive acceleration alone does not fix direction, a zero velocity does not pin the location, and equal values carry no special meaning.
- Water is draining from a conical tank so that the height of the water decreases at 2 centimeters per minute. The differentiation of the volume relationship with respect to time is used to find how fast a different quantity changes. This describes which type of problem?
- An optimization problem
- A linear approximation problem
- A related rates problem
- An average value problem
Correct answer: A related rates problem
This is a related rates problem because a known time rate, the height decreasing at 2 centimeters per minute, is connected through an equation to an unknown time rate such as the volume's rate of change. Differentiating the volume relationship with respect to time links the two rates. Optimization finds extrema, linear approximation estimates values with a tangent line, and average value involves an integral, none of which match.
- The side of a square is increasing at a rate of 4 inches per second. How fast is the area increasing when the side length is 10 inches?
- 40 square inches per second
- 80 square inches per second
- 100 square inches per second
- 20 square inches per second
Correct answer: 80 square inches per second
The area increases at 80 square inches per second. Differentiating the area A = s² with respect to time gives (dA)/dt = 2s · ds/dt. Substituting s = 10 and ds/dt = 4 produces 2 · 10 · 4 = 80.
- A particle's position is given by s(t) = 2t³ - 9t² + 12t. What is the particle's acceleration at time t = 1?
Correct answer: -6
The acceleration at t = 1 is -6. Velocity is the first derivative, v(t) = 6t² - 18t + 12, and acceleration is its derivative, a(t) = 12t - 18. Substituting t = 1 gives 12 - 18 = -6.
- Evaluate lim x → 0 (1 - cos x)/x² using L'Hopital's Rule.
Correct answer: 1/2
The limit equals 1/2. The form is 0/0, so one application of L'Hopital's Rule gives (sin x)/(2x), still 0/0, requiring a second application to get (cos x)/2. Evaluating (cos x)/2 at x = 0 yields 1/2.
- Using the local linearization of f(x) = ln x centered at x = 1, estimate ln 1.1.
Correct answer: 0.1
The estimate is 0.1. At x = 1, f(1) = 0 and f'(x) = 1/x gives f'(1) = 1. The linearization is 0 + 1 · 0.1 = 0.1.
- A 26-foot ladder rests against a wall, and the top slides down at 3 feet per second. When the top is 10 feet above the ground, how fast is the base moving away from the wall?
- 3/8 of a foot per second
- 10/24 of a foot per second
- 8/5 of a foot per second
- 5/8 of a foot per second
Correct answer: 5/8 of a foot per second
The base moves away at 5/8 of a foot per second. With x² + y² = 676, when y = 10 the base x = 24. Differentiating gives 2x · dx/dt + 2y · dy/dt = 0, so dx/dt = -(y · dy/dt)/x = -(10 · (-3))/24 = 30/24 = 5/8.
- Evaluate lim x → 0 (tan x)/x using L'Hopital's Rule.
Correct answer: 1
The limit equals 1. Direct substitution gives the indeterminate form 0/0, so L'Hopital's Rule allows differentiating to get (sec² x)/1. Evaluating sec² x at x = 0 gives sec² 0 = 1² = 1.
- A function f satisfies f(4) = 7 and f'(4) = 3. Use a tangent-line approximation to estimate f(4.2).
Correct answer: 7.6
The estimate is 7.6. The linear approximation is f(4) + f'(4) · Δ x = 7 + 3 · 0.2. That gives 7 + 0.6 = 7.6.
- A particle moves along a line with velocity v(t) = t² - 5t + 6 for t ≥ 0. On which interval is the particle moving to the left?
- From t = 0 to t = 2
- From t = 2 to t = 3
- From t = 3 onward
- From t = 0 to t = 3
Correct answer: From t = 2 to t = 3
The particle moves left from t = 2 to t = 3. Factoring gives v(t) = (t - 2)(t - 3), which is zero at t = 2 and t = 3. Between these roots the product is negative, meaning velocity is negative and the particle moves left, while outside that interval velocity is positive.
- When using a tangent-line approximation for a function that is concave up near the point of tangency, how does the estimate compare to the true function value?
- The estimate equals the true value exactly
- The estimate is an underestimate of the true value
- The estimate is an overestimate of the true value
- The comparison cannot be determined from concavity
Correct answer: The estimate is an underestimate of the true value
For a function that is concave up near the point of tangency, the tangent line lies below the curve, so the linear approximation underestimates the true value. Concavity describes how the graph bends away from its tangent line, and upward bending keeps the curve above the line. An exact match would require a linear function, and concavity does in fact determine the direction of the error.
- Sand falls onto a conical pile so that its volume increases at 12 cubic feet per minute, and at a certain instant the rate relationship reduces to 12 = 9π · dr/dt. What is dr/dt at that instant?
- 4/(3π) feet per minute
- 3/(4π) feet per minute
- 12/π feet per minute
- 9/π feet per minute
Correct answer: 4/(3π) feet per minute
The radius increases at 4/(3π) feet per minute. Starting from 12 = 9π · dr/dt, dividing both sides by 9π isolates dr/dt = 12/(9π). Reducing 12/9 to 4/3 gives dr/dt = 4/(3π).
- Evaluate lim x → 0 (x - sin x)/x³ using L'Hopital's Rule.
Correct answer: 1/6
The limit equals 1/6. The form is 0/0, so applying L'Hopital's Rule gives (1 - cos x)/(3x²), still 0/0; a second application gives (sin x)/(6x), again 0/0; a third gives (cos x)/6. Evaluating (cos x)/6 at x = 0 yields 1/6.
- A particle moving along a line has velocity v(t) = 3t² - 12 and is at position s = 5 when t = 0. Without integrating, what does evaluating the velocity at t = 2 tell you about the particle's motion at that instant?
- The particle is moving and instantaneously not accelerating
- The particle is at position 5 again
- The particle has returned to its starting point
- The particle is momentarily at rest, since the velocity equals zero there
Correct answer: The particle is momentarily at rest, since the velocity equals zero there
At t = 2 the particle is momentarily at rest because v(2) = 3 · 4 - 12 = 12 - 12 = 0, and zero velocity means no instantaneous motion. The acceleration there is a(t) = 6t, giving 12, so it is in fact accelerating. Velocity alone gives no information about the position value, so the particle's location cannot be concluded to be 5 or the start.
- What two hypotheses must a function f satisfy on the interval from a to b for the Mean Value Theorem to apply?
- f must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b)
- f must be differentiable on the closed interval [a, b] and have f(a) = f(b)
- f must be increasing on the interval from a to b and continuous at the endpoints
- f must be continuous on the open interval (a, b) and have a maximum on that interval
Correct answer: f must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b)
The correct requirement is that f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). The Mean Value Theorem then guarantees at least one interior point c where the instantaneous rate of change f'(c) equals the average rate of change, (f(b) - f(a))/(b - a). Equal endpoint values describe the special case Rolle's Theorem, not the general hypotheses, and being increasing is not required.
- The function f(x) = x² is defined on the closed interval [1, 4]. According to the Mean Value Theorem, there exists a value c in the open interval where f'(c) equals the average rate of change. What is that value of c?
Correct answer: c = 5/2
The value guaranteed by the theorem is c = 5/2. The average rate of change is (f(4) - f(1))/(4 - 1) = (16 - 1)/3 = 5. Setting f'(c) = 2c = 5 gives c = 5/2 = 2.5, which lies inside the open interval (1, 4).
- The Extreme Value Theorem guarantees that a function attains both an absolute maximum and an absolute minimum under which condition?
- When f is differentiable on an open interval
- When f is increasing throughout its domain
- When f is continuous on a closed, bounded interval
- When f has at least one critical point
Correct answer: When f is continuous on a closed, bounded interval
The guarantee holds when f is continuous on a closed, bounded interval. The Extreme Value Theorem states that a function continuous on such an interval must achieve both an absolute maximum and an absolute minimum value somewhere on that interval. Differentiability on an open interval, being increasing, or merely having a critical point does not by itself ensure that absolute extrema are attained.
- Find all critical points of the function f(x) = x³ - 3x² - 9x + 5.
- x = 0 and x = 3
- x = 1 and x = -3
- x = -3 and x = 9
- x = -1 and x = 3
Correct answer: x = -1 and x = 3
The critical points occur at x = -1 and x = 3. Differentiating gives f'(x) = 3x² - 6x - 9, which factors as 3(x - 3)(x + 1). Setting this equal to zero yields x = 3 and x = -1, the only values where the derivative is zero, and the derivative exists everywhere.
- A differentiable function f has f'(x) = (x - 2)(x + 4). Using the First Derivative Test, classify the critical point at x = 2.
- A relative minimum, because f' changes from negative to positive at x = 2
- A relative maximum, because f' changes from positive to negative at x = 2
- Neither a maximum nor a minimum, because f' does not change sign
- A relative maximum, because f' is zero at x = 2
Correct answer: A relative minimum, because f' changes from negative to positive at x = 2
The point at x = 2 is a relative minimum because f' changes from negative to positive there. For x slightly less than 2, the factor (x - 2) is negative and (x + 4) is positive, so f' is negative; for x slightly greater than 2, both factors are positive, so f' is positive. A change from decreasing to increasing identifies a relative minimum.
- Using the Second Derivative Test, classify the critical point of f(x) = x² - 6x + 1 at x = 3.
- A relative maximum, because the second derivative is positive there
- A relative minimum, because the second derivative is positive there
- A relative maximum, because the second derivative is negative there
- A point of inflection, because the second derivative is zero there
Correct answer: A relative minimum, because the second derivative is positive there
The point is a relative minimum because the second derivative is positive there. The first derivative f'(x) = 2x - 6 is zero at x = 3, and the second derivative f''(x) = 2 is positive. A positive second derivative at a critical point means the graph is concave up, identifying a relative minimum.
- For the function f(x) = x³ - 6x² + 5, on what interval is the graph concave down?
- For x > 2
- For x > 0
- For x < 2
- For all real x
Correct answer: For x < 2
The graph is concave down for x < 2. The second derivative is f''(x) = 6x - 12, which is negative when x < 2. A negative second derivative indicates concave-down behavior, so the curve bends downward on that interval and switches to concave up beyond x = 2.
- A farmer wants to enclose a rectangular field bordering a straight river, with no fence needed along the river, using 200 meters of fencing for the other three sides. What dimensions maximize the enclosed area?
- 50 meters parallel to the river and 75 meters for each perpendicular side
- 200 meters parallel to the river and 0 meters perpendicular
- 66 meters parallel to the river and 67 meters for each perpendicular side
- 100 meters parallel to the river and 50 meters for each perpendicular side
Correct answer: 100 meters parallel to the river and 50 meters for each perpendicular side
The maximizing dimensions are 100 meters parallel to the river and 50 meters for each side perpendicular to it. If x is each perpendicular side, the side parallel to the river is 200 - 2x, so area A = x(200 - 2x). Setting A' = 200 - 4x to zero gives x = 50, making the parallel side 100 meters and the maximum area 5000 square meters.
- Find the absolute maximum value of f(x) = x³ - 3x on the closed interval [-2, 2].
Correct answer: 2
The absolute maximum value is 2. The derivative f'(x) = 3x² - 3 is zero at x = -1 and x = 1. Evaluating f at the critical points and endpoints gives f(-2) = -2, f(-1) = 2, f(1) = -2, and f(2) = 2. The largest of these output values is 2.
- The function f(x) = x⁴ - 4x³ has a point of inflection where the concavity changes. At which x-value does an inflection point occur, other than x = 0?
Correct answer: x = 2
An inflection point occurs at x = 2. The second derivative is f''(x) = 12x² - 24x, which factors as 12x(x - 2). It equals zero at x = 0 and x = 2, and the concavity actually changes sign at both. Other than x = 0, the inflection point is at x = 2.
- A continuous function f satisfies f(0) = 4 and f(6) = 16. What does the Mean Value Theorem guarantee about the rate of change of f, assuming f is differentiable on the open interval?
- There is some point c where f'(c) = 0
- The function f is increasing on the entire interval
- There is some point c where f'(c) = 2
- There is some point c where f(c) = 10
Correct answer: There is some point c where f'(c) = 2
The theorem guarantees a point c where f'(c) = 2. The average rate of change is (16 - 4)/(6 - 0), which equals 12/6, or 2. The Mean Value Theorem ensures at least one interior point where the instantaneous rate of change equals this average. A guaranteed function value such as f(c) = 10 is a statement of the Intermediate Value Theorem, not this one.
- A function f is differentiable everywhere, and its first derivative f' is positive on the interval (-∞, 1) and negative on the interval (1, ∞). What can be concluded about f at x = 1?
- f has a relative minimum at x = 1
- f has an inflection point at x = 1
- f is not defined at x = 1
- f has a relative maximum at x = 1
Correct answer: f has a relative maximum at x = 1
The function has a relative maximum at x = 1. By the First Derivative Test, when f' changes from positive to negative at a critical point, the function changes from increasing to decreasing, producing a relative maximum. A change in the opposite direction would give a minimum, and concavity information would be needed to identify an inflection point.
- An open-top box is made from a square sheet of cardboard 12 inches on each side by cutting equal squares of side x from the corners and folding up the sides. What value of x maximizes the volume of the box?
Correct answer: x = 2
The volume is maximized when x = 2 inches. The volume is V = x(12 - 2x)². Differentiating and setting V' = 0 leads to 12(6 - x)(2 - x) = 0, giving x = 2 or x = 6. Since x = 6 collapses the base to zero, the maximum occurs at x = 2, yielding a volume of 128 cubic inches.
- Why can a critical point where f' = 0 fail to be a relative extremum?
- Because the function must be discontinuous at every critical point
- Because the first derivative may not change sign as x passes through the point
- Because critical points only occur at the endpoints of an interval
- Because a zero derivative always indicates an inflection point
Correct answer: Because the first derivative may not change sign as x passes through the point
A critical point fails to be an extremum when the first derivative does not change sign as x passes through it. For example, f(x) = x³ has f' = 3x², which is zero at x = 0 but stays nonnegative on both sides, so the function keeps increasing and no extremum occurs. A sign change in f' is what distinguishes a genuine relative maximum or minimum.
- The second derivative of a function f is given by f''(x) = 6x - 18. On what interval is the graph of f concave up?
- For x < 3
- For x > 18
- For x > 3
- For all real x
Correct answer: For x > 3
The graph is concave up for x > 3. Setting f''(x) = 6x - 18 > 0 gives x > 3, where the second derivative is positive. A positive second derivative means the curve bends upward, so concavity is upward to the right of x = 3.
- The Second Derivative Test is inconclusive at a critical point when the second derivative equals zero there. In that situation, which approach can still determine whether the point is a relative extremum?
- Conclude immediately that the point is an inflection point
- Conclude that no extremum can exist at the point
- Evaluate the third derivative, which always settles the classification
- Apply the First Derivative Test by checking the sign of f' on both sides
Correct answer: Apply the First Derivative Test by checking the sign of f' on both sides
When the Second Derivative Test is inconclusive, the First Derivative Test resolves the question by checking the sign of f' on both sides of the critical point. A change from positive to negative indicates a relative maximum and negative to positive indicates a relative minimum; no sign change means neither. The vanishing of the second derivative does not by itself prove an inflection point or rule out an extremum.
- Find the absolute minimum value of f(x) = x² - 4x + 7 on the closed interval [0, 5].
Correct answer: 3
The absolute minimum value is 3. The derivative f'(x) = 2x - 4 is zero at x = 2, the only critical point in the interval. Evaluating gives f(0) = 7, f(2) = 3, and f(5) = 12, so the smallest output value over the closed interval is 3, occurring at x = 2.
- A right Riemann sum is used to approximate the area under a continuous, increasing function on a closed interval using equal-width rectangles. Compared with the exact area under the curve, this approximation will be:
- An overestimate, because each rectangle's right endpoint gives the largest function value on its subinterval
- An underestimate, because each rectangle lies entirely below the curve
- Exactly equal, because the rectangle widths are equal
- Sometimes larger and sometimes smaller, depending on the number of rectangles
Correct answer: An overestimate, because each rectangle's right endpoint gives the largest function value on its subinterval
The approximation is an overestimate because the function is increasing. For an increasing function, the right endpoint of each subinterval produces the greatest function value on that subinterval, so every rectangle's height exceeds the curve over most of its width, making the total area too large. A left Riemann sum would underestimate for the same increasing function.
- The Fundamental Theorem of Calculus states that if f is continuous on the closed interval [a, b] and F is an antiderivative of f, then ∫ from a to b of f(x) dx equals which of the following?
- F(b) + F(a)
- F(b) - F(a)
- f'(b) - f'(a)
- F'(b) - F'(a)
Correct answer: F(b) - F(a)
The correct evaluation is F(b) - F(a). The second part of the Fundamental Theorem of Calculus says that to evaluate a definite integral of f, you find any antiderivative F and compute its value at the upper limit minus its value at the lower limit. The other expressions misuse addition or apply the rule to derivatives rather than to the antiderivative.
- Evaluate the definite integral ∫ from 1 to 3 of 2x dx.
Correct answer: 8
The value of the integral is 8. An antiderivative of 2x is x², so by the Fundamental Theorem of Calculus the integral equals 3² - 1², which is 9 - 1, or 8. Forgetting to subtract the lower-limit value or mishandling the antiderivative produces the other choices.
- Use u-substitution to evaluate the indefinite integral ∫ 2x(x² + 1)⁴ dx.
- (x² + 1)⁵ + C
- 1/4(x² + 1)⁵ + C
- 5(x² + 1)⁵ + C
- 1/5(x² + 1)⁵ + C
Correct answer: 1/5(x² + 1)⁵ + C
The antiderivative is 1/5(x² + 1)⁵ + C. Letting u = x² + 1 gives du = 2x dx, so the integral becomes ∫ u⁴ du, which is 1/5u⁵. Substituting back gives the stated result; the other options drop the 1/5 factor or use the wrong power-rule coefficient.
- Let g be the accumulation function defined by g(x) = ∫ from 0 to x of f(t) dt, where f is continuous. According to the first part of the Fundamental Theorem of Calculus, what is g'(x)?
- f(x)
- f'(x)
- ∫ from 0 to x of f(t) dt
- f(x) - f(0)
Correct answer: f(x)
The derivative g'(x) = f(x). The first part of the Fundamental Theorem of Calculus states that the derivative of an accumulation function whose upper limit is x is simply the integrand evaluated at x. This is why differentiation and integration are inverse processes; the other choices confuse the integrand with its derivative or with the accumulated value itself.
- A car's velocity is recorded at several times during a trip, and a midpoint Riemann sum is used to approximate the integral of the velocity over the time interval. In context, what quantity does this Riemann sum approximate?
- The car's acceleration at the midpoint of the interval
- The car's average velocity over the interval
- The total distance traveled by the car over the time interval
- The maximum velocity reached during the trip
Correct answer: The total distance traveled by the car over the time interval
The Riemann sum approximates the total distance traveled. Integrating velocity over a time interval accumulates the change in position, which for nonnegative velocity is the total distance. A Riemann sum estimates that definite integral, not acceleration (a rate of change of velocity), average velocity, or a single maximum value.
- Find the derivative with respect to x of the accumulation function H(x) = ∫ from 2 to x of √(1 + t³) dt.
- √(1 + t³)
- 1/2(1 + x³)^(-1/2)
- √(1 + x³) - √9
- √(1 + x³)
Correct answer: √(1 + x³)
The derivative is √(1 + x³). By the first part of the Fundamental Theorem of Calculus, when the upper limit is x and the lower limit is a constant, the derivative of the accumulation function is the integrand with t replaced by x. The constant lower limit 2 does not affect the derivative, so no extra subtracted term appears.
- Evaluate the definite integral ∫ from 0 to π/2 of cos(x) dx.
Correct answer: 1
The value of the integral is 1. An antiderivative of cos x is sin x, so the integral equals sin(π/2) - sin(0), which is 1 - 0, or 1. Confusing the antiderivative of cosine with -sin x, or mis-evaluating the sine values, leads to the incorrect choices.
- Use an appropriate u-substitution to evaluate the definite integral ∫ from 0 to √(π/2) of xcos(x²) dx.
Correct answer: 1/2
The value is 1/2. Letting u = x² gives du = 2x dx, so x dx = 1/2 du, and the limits become u from 0 to π/2. The integral becomes 1/2∫ cos u du, which is 1/2sin u evaluated from 0 to π/2, equal to 1/2(1 - 0), or 1/2.
- A left Riemann sum with four equal subintervals is used to approximate the integral of a continuous, decreasing, positive function on a closed interval. How does this approximation compare with the exact value of the integral?
- It underestimates the integral
- It equals the integral exactly
- It overestimates the integral
- It cannot be compared without knowing the function values
Correct answer: It overestimates the integral
The left Riemann sum overestimates the integral. For a decreasing function, the left endpoint of each subinterval gives the largest function value there, so each rectangle is taller than the curve over most of its width, producing too much area. A right Riemann sum would underestimate the same decreasing function.
- The temperature of an object changes at a rate of r(t) degrees per hour. If the object's temperature at time t = 0 is 60 degrees, which expression gives the temperature at time t = 5 using accumulation of change?
- ∫ from 0 to 5 of r(t) dt
- 60 + r(5) - r(0)
- 60 · ∫ from 0 to 5 of r(t) dt
- 60 + ∫ from 0 to 5 of r(t) dt
Correct answer: 60 + ∫ from 0 to 5 of r(t) dt
The temperature equals 60 + ∫ from 0 to 5 of r(t) dt. The definite integral of a rate of change gives the net change in the quantity over the interval, so adding that accumulated change to the initial value of 60 degrees produces the final temperature. Using r-values directly or omitting the initial condition gives the incorrect choices.
- Find the indefinite integral ∫ 1/x dx.
- ln|x| + C
- -1/x² + C
- x⁻¹ + C
- ln(x²) + C
Correct answer: ln|x| + C
The antiderivative is ln|x| + C. This is the standard antiderivative for the reciprocal function; the absolute value extends the result to negative x-values. The power rule for antiderivatives fails for the exponent -1, which is why this special form is needed rather than the choices based on power-rule patterns.
- Use u-substitution to evaluate the indefinite integral ∫ e³ˣ dx.
- 3e³ˣ + C
- e³ˣ + C
- 1/3e³ˣ + C
- 1/3eˣ + C
Correct answer: 1/3e³ˣ + C
The antiderivative is 1/3e³ˣ + C. Letting u = 3x gives du = 3 dx, so dx = 1/3 du, and the integral becomes 1/3∫ e^u du. The result keeps the exponent 3x and includes the 1/3 factor that the other choices either omit or invert.
- The function g is defined by g(x) = ∫ from 1 to x of f(t) dt, where f is a continuous function whose graph lies entirely above the t-axis for t > 1. What can be concluded about g on the interval x > 1?
- g is decreasing, because the lower limit is fixed
- g is constant, because integration removes the variable
- g is negative, because integration reverses sign
- g is increasing, because g'(x) = f(x) is positive there
Correct answer: g is increasing, because g'(x) = f(x) is positive there
The function g is increasing on that interval. By the Fundamental Theorem of Calculus, g'(x) = f(x), and since f is positive for t > 1, the derivative of g is positive, so g increases. The accumulated area grows as the upper limit advances; the fixed lower limit does not make g decrease or stay constant.
- Evaluate the definite integral ∫ from 0 to 2 of (6x² - 4x) dx.
Correct answer: 8
The value of the integral is 8. An antiderivative of 6x² - 4x is 2x³ - 2x². Evaluating at the upper limit gives 2 · 8 - 2 · 4, which is 16 - 8, or 8; at the lower limit the antiderivative is 0. Subtracting gives 8 - 0, or 8.
- A midpoint Riemann sum with two equal subintervals approximates the integral of f over the interval [0, 4], where f(1) = 5 and f(3) = 9. What is the value of this approximation?
Correct answer: 28
The approximation is 28. With two subintervals over a width-4 interval, each subinterval has width 2, and the midpoints are x = 1 and x = 3. The midpoint sum is 2 · f(1) + 2 · f(3), which is 2 · 5 + 2 · 9, equal to 10 + 18, or 28. Forgetting to multiply by the subinterval width gives the smaller incorrect values.
- Use u-substitution to evaluate the indefinite integral ∫ x²/(x³ + 1) dx.
- ln|x³ + 1| + C
- 3ln|x³ + 1| + C
- 1/3ln|x³ + 1| + C
- 1/3(x³ + 1) + C
Correct answer: 1/3ln|x³ + 1| + C
The antiderivative is 1/3ln|x³ + 1| + C. Letting u = x³ + 1 gives du = 3x² dx, so x² dx = 1/3 du, and the integral becomes 1/3∫ 1/u du. The antiderivative of 1/u is ln|u|, producing the stated result.
- Which of the following is the defining characteristic of a separable first-order differential equation?
- It can be written so that all terms involving y are on one side and all terms involving x are on the other side before integrating.
- It has constant coefficients and a forcing term equal to zero.
- Its solution must always be an exponential function of x.
- It can only be solved using a slope field rather than by integration.
Correct answer: It can be written so that all terms involving y are on one side and all terms involving x are on the other side before integrating.
A separable differential equation is one that can be rewritten so all y-terms (including dy) are on one side and all x-terms (including dx) are on the other, allowing each side to be integrated independently. Constant coefficients describe linear equations, not separability; the solution need not be exponential; and a slope field is a graphical tool, not a defining property.
- A quantity P changes over time t according to (dP)/dt = kP, where k is a positive constant. Which type of model does this differential equation represent?
- Linear decay at a constant rate
- Exponential growth proportional to the current amount
- Logistic growth approaching a fixed carrying capacity
- Periodic oscillation about an equilibrium
Correct answer: Exponential growth proportional to the current amount
The equation (dP)/dt = kP with k positive describes exponential growth, because the rate of change is proportional to the current amount P, yielding solutions of the form P = P₀ e^(kt). Linear decay would have a constant negative rate; logistic growth includes a carrying-capacity factor; and periodic oscillation would require a trigonometric form.
- Solve the separable differential equation dy/dx = 6x² y given the initial condition y(0) = 5.
- y = 5 e^(6x³)
- y = 5 e^(3x²)
- y = 2x³ + 5
- y = 5 e^(2x³)
Correct answer: y = 5 e^(2x³)
The solution is y = 5 e^(2x³). Separating gives 1/y dy = 6x² dx; integrating yields ln|y| = 2x³ + C since the antiderivative of 6x² is 2x³. Exponentiating gives y = A e^(2x³), and applying y(0) = 5 makes A = 5. Keeping the coefficient 6 in the exponent or failing to integrate the power correctly produces the other forms.
- A slope field for a differential equation dy/dx = f(x, y) shows short line segments at selected points in the plane. What does the slope of each segment represent?
- The exact value of the solution y at that point
- The area under the solution curve up to that point
- The value of dy/dx, the slope of a solution curve passing through that point
- The second derivative of the solution at that point
Correct answer: The value of dy/dx, the slope of a solution curve passing through that point
Each segment in a slope field shows the value of dy/dx at that point, which is the slope of any solution curve passing through it. It does not give the solution's y-value, an accumulated area, or the second derivative; it visualizes the direction a solution travels at each location.
- The differential equation dy/dx = x/y has a slope field. Along which set of points are the segments horizontal?
- Along the line y = x
- Along the y-axis, where x = 0
- Along the x-axis, where y = 0
- Along the line y = -x
Correct answer: Along the y-axis, where x = 0
The segments are horizontal along the y-axis where x = 0, because dy/dx = x/y equals 0 exactly when the numerator x is 0 (with y ≠ 0). A zero slope means horizontal segments. The x-axis (y = 0) would make the expression undefined, and the lines y = x or y = -x give slopes of 1 or -1, not 0.
- A radioactive sample decays according to (dA)/dt = -0.04 A, where A is measured in grams and t in years. If the initial amount is 200 grams, which expression gives the amount remaining after t years?
- A = 200 - 0.04t
- A = 200 e^(0.04t)
- A = 200 e^(-0.04t)
- A = 0.04 e^(-200t)
Correct answer: A = 200 e^(-0.04t)
The amount is A = 200 e^(-0.04t). The equation (dA)/dt = -0.04 A is exponential decay with rate constant -0.04, so the solution is A = A₀ e^(-0.04t) with A₀ = 200. A linear form ignores the proportional rate, a positive exponent would describe growth, and swapping the constants misuses the initial value.
- Find the particular solution to dy/dx = (2x)/y that satisfies y(0) = 3, expressed as a function of x.
- y = √(2x² + 3)
- y = √(2x² + 9)
- y = x² + 3
- y = √(x² + 9)
Correct answer: y = √(2x² + 9)
The particular solution is y = √(2x² + 9). Separating gives y dy = 2x dx; integrating yields 1/2 y² = x² + C, so y² = 2x² + 2C. Applying y(0) = 3 gives 9 = 2C, so y² = 2x² + 9 and y = √(2x² + 9), taking the positive root to match the initial value.
- A population grows exponentially and doubles every 10 years. If the model is P = P₀ e^(kt) with t in years, which equation must the growth constant k satisfy?
- k = 10/(ln 2)
- k = (ln 10)/2
- k = 2/10
- k = (ln 2)/10
Correct answer: k = (ln 2)/10
The growth constant satisfies k = (ln 2)/10. Doubling means P = 2 P₀ when t = 10, so 2 = e^(10k); taking the natural log gives ln 2 = 10k, hence k = (ln 2)/10. The other forms invert the relationship or replace the doubling factor incorrectly.
- Two students each sketch a solution curve on the slope field for dy/dx = y. One starts at a point where y is positive and the other where y is negative. How do their solution curves behave as x increases?
- Both curves flatten toward the line y = 0 as x increases.
- Both curves move away from y = 0, the positive one rising and the negative one falling, since the slope grows in magnitude with |y|.
- Both curves oscillate up and down because the slope alternates sign.
- Both curves become vertical because the slope is undefined where y = 0.
Correct answer: Both curves move away from y = 0, the positive one rising and the negative one falling, since the slope grows in magnitude with |y|.
Both curves move away from y = 0—the positive one rising and the negative one falling—because dy/dx = y makes the slope positive when y is positive and negative when y is negative, with magnitude increasing as |y| grows. The line y = 0 is an unstable equilibrium, so curves diverge from it rather than flattening toward it, oscillating, or going vertical.
- When solving the separable equation dy/dx = (cos x)/(2y) for the general solution, what is the result after integrating both sides?
- y = sin x + C
- y² = sin x + C
- 2y = cos x + C
- y² = -cos x + C
Correct answer: y² = sin x + C
After integrating, the general solution satisfies y² = sin x + C. Separating gives 2y dy = cos x dx; integrating the left side yields y² and the right side yields sin x plus a constant, so y² = sin x + C. The other options drop the squared term, mishandle the factor of 2, or use the antiderivative of sin x instead of cos x.
- To find the area of the region bounded between the curves y = f(x) on top and y = g(x) on the bottom from x = a to x = b, which definite integral should be evaluated?
- ∫ from a to b of (f(x) - g(x)) dx
- ∫ from a to b of (f(x) + g(x)) dx
- ∫ from a to b of (g(x) - f(x)) dx
- ∫ from a to b of f(x) g(x) dx
Correct answer: ∫ from a to b of (f(x) - g(x)) dx
The correct setup is ∫ from a to b of (f(x) - g(x)) dx, the top curve minus the bottom curve. Each thin vertical slice of the region has height equal to the upper function value minus the lower function value, and integrating that height across the interval accumulates the total area. Adding the curves or multiplying them does not represent the height of a slice, and reversing to g(x) - f(x) would give a negative of the area when f is on top.
- Find the area of the region enclosed between the curves y = x² and y = 2x.
Correct answer: 4/3
The enclosed area is 4/3. The curves intersect where x² = 2x, which gives x = 0 and x = 2, and on that interval the line y = 2x lies above the parabola y = x². Integrating 2x - x² from 0 to 2 gives [x² - x³/3]₀², which is 4 - 8/3 = 4/3.
- A solid is generated by revolving the region under y = √x, from x = 0 to x = 4, about the x-axis. Using the disk method, which integral gives the volume?
- π ∫ from 0 to 4 of x dx
- π ∫ from 0 to 4 of √x dx
- 2π ∫ from 0 to 4 of x dx
- π ∫ from 0 to 4 of x² dx
Correct answer: π ∫ from 0 to 4 of x dx
The volume is π ∫ from 0 to 4 of x dx. With the disk method, each disk has radius equal to the function value, √x, so its area is π (√x)², and squaring √x gives x. Integrating π x represents stacking the disk areas along the axis of revolution.
- The region bounded by y = x², the x-axis, and the line x = 2 is revolved about the y-axis using the washer method with respect to y. Across the interval from y = 0 to y = 4, what are the outer and inner radii of a typical washer?
- Outer radius √y and inner radius 2
- Outer radius y and inner radius 2
- Outer radius 4 and inner radius y²
- Outer radius 2 and inner radius √y
Correct answer: Outer radius 2 and inner radius √y
The outer radius is 2 and the inner radius is √y. Revolving about the y-axis, the outer edge of the region is the vertical line x = 2, giving a constant outer radius of 2, while the inner edge is the curve x = √y, obtained by solving y = x² for x. The washer's area is π (2² - (√y)²), here π (4 - y).
- A solid has a base bounded by the circle x² + y² = 9, and every cross section perpendicular to the x-axis is a square. If the side length of each square equals the vertical extent of the base, which integral gives the volume?
- ∫ from -3 to 3 of 4(9 - x²) dx
- ∫ from -3 to 3 of (9 - x²) dx
- π ∫ from -3 to 3 of (9 - x²) dx
- ∫ from -3 to 3 of 2√(9 - x²) dx
Correct answer: ∫ from -3 to 3 of 4(9 - x²) dx
The volume is ∫ from -3 to 3 of 4(9 - x²) dx. At a given x the base extends from -√(9 - x²) to +√(9 - x²), so the square's side length is 2√(9 - x²). Squaring that side gives 4(9 - x²), and there is no factor of π because the cross sections are squares, not circles.
- Find the volume of the solid formed when the region bounded by y = x, the x-axis, and x = 3 is revolved about the x-axis.
Correct answer: 9π
The volume is 9π. Each disk has radius equal to y = x, so its area is π x², and the volume is π ∫ from 0 to 3 of x² dx. That integral equals [x³/3]₀³, which is 27/3 = 9, giving 9π.
- Find the area of the region between the curves y = sin x and y = cos x from x = π/4 to x = (5π)/4.
Correct answer: 2√2
The area is 2√2. On the interval from π/4 to (5π)/4 the sine curve lies above the cosine curve, so the area is ∫ (sin x - cos x) dx. The antiderivative is -cos x - sin x, and evaluating it from π/4 to (5π)/4 produces 2√2.
- When a region between two curves is revolved about a horizontal axis and there is a gap between the inner curve and the axis of revolution, why is the washer method used instead of the disk method?
- The washer method applies only to cross sections that are triangles
- The disk method cannot handle horizontal axes of revolution
- The gap creates a hollow center, so the volume of the inner region must be subtracted from the outer region
- The washer method avoids the use of π in the integrand
Correct answer: The gap creates a hollow center, so the volume of the inner region must be subtracted from the outer region
The washer method is used because the gap between the inner curve and the axis leaves a hollow core that must be removed. Each cross section is a ring, or washer, whose area is π (R² - r²), the outer radius squared minus the inner radius squared, subtracting the missing inner disk. The disk method assumes a solid cross section with no hole, so it would overcount the volume when a gap is present.
- The region bounded by y = x² and y = 4 is revolved about the x-axis. Using the washer method, which integral gives the volume?
- π ∫ from -2 to 2 of (4 - x²) dx
- π ∫ from -2 to 2 of (x⁴ - 16) dx
- π ∫ from 0 to 4 of (16 - x⁴) dx
- π ∫ from -2 to 2 of (16 - x⁴) dx
Correct answer: π ∫ from -2 to 2 of (16 - x⁴) dx
The volume is π ∫ from -2 to 2 of (16 - x⁴) dx. Revolving about the x-axis, the outer radius is the line y = 4 and the inner radius is the parabola y = x², since the curves meet at x = -2 and x = 2. The washer area is π (4² - (x²)²), which is π (16 - x⁴).
- Find the volume of the solid generated by revolving the region bounded by y = x², the y-axis, and y = 1 about the y-axis.
Correct answer: π/2
The volume is π/2. Revolving about the y-axis, integrate with respect to y, where each disk has radius x = √y from the curve y = x². The volume is π ∫ from 0 to 1 of (√y)² dy, which is π ∫ from 0 to 1 of y dy = π [y²/2]₀¹ = π/2.
- A solid has a base bounded by y = x and y = x², and every cross section perpendicular to the x-axis is a semicircle whose diameter lies in the base. Which expression represents the area of a cross section at a given x?
- π/2(x - x²)
- π (x - x²)²
- π/8(x - x²)²
- (x - x²)²
Correct answer: π/8(x - x²)²
The cross-sectional area is π/8(x - x²)². The diameter equals the height of the base, x - x², so the radius is (x - x²)/2. A semicircle's area is 1/2 π r² = 1/2 π ((x - x²)/2)², simplifying to π/8(x - x²)².
- A student sets up the area between y = x² and y = x + 2 as ∫ from -1 to 2 of (x² - x - 2) dx and obtains a negative number. What mistake explains the negative result?
- The limits of integration are reversed
- The student subtracted the upper curve from the lower curve instead of the other way around
- The two curves do not actually intersect
- Area integrals are always negative and the answer should be left as is
Correct answer: The student subtracted the upper curve from the lower curve instead of the other way around
The mistake is subtracting the top curve from the bottom curve in the wrong order. On the interval from -1 to 2 the line y = x + 2 lies above the parabola y = x², so the integrand should be x + 2 - x², not x² - x - 2. Reversing the subtraction flips the sign, producing a negative value whose absolute value is the true area of 9/2.
- Using a table of values, a student finds that f(x) gets closer and closer to 7 as x approaches 3 from both sides, even though f(3) = 2. What does this table-based reasoning suggest about lim x → 3 f(x)?
- lim x → 3 f(x) = 4.5
- lim x → 3 f(x) = 2
- The limit does not exist because f(3) is not 7
- lim x → 3 f(x) = 7
Correct answer: lim x → 3 f(x) = 7
The limit lim x → 3 f(x) = 7. A limit describes the value the function approaches as x nears the input, which the table shows is 7 from both sides. The actual function value f(3) = 2 is irrelevant to the limit, so 2 and the average 4.5 are wrong, and the disagreement between limit and function value does not make the limit fail to exist.
- Evaluate lim x → 3 (x² - x - 6)/(x - 3).
- 0
- 5
- 3
- The limit does not exist
Correct answer: 5
The answer is 5. Factor the numerator as (x - 3)(x + 2), then cancel the common (x - 3) factor to get x + 2. Substituting x = 3 into x + 2 gives 5. The expression has a removable hole at x = 3 rather than a nonexistent limit, so 0, 3, and 'does not exist' are incorrect.
- Evaluate lim x → 4 (1/x - 1/4)/(x - 4).
Correct answer: -1/16
The answer is -1/16. Combine the numerator over a common denominator to get (4 - x)/(4x), so the whole expression becomes (4 - x)/(4x(x - 4)). Since 4 - x = -(x - 4), this simplifies to -1/(4x). Substituting x = 4 gives -1/16.
- Evaluate lim x → 0 (tan x)/x.
- ∞
- 0
- 1
- The limit does not exist
Correct answer: 1
The answer is 1. Write (tan x)/x as (sin x)/x · 1/(cos x). As x → 0, (sin x)/x → 1 and 1/(cos x) → 1, so the product approaches 1 · 1 = 1. The values 0, ∞, and 'does not exist' do not match this standard trigonometric limit.
- Evaluate lim x → 0 (1 - cos x)/x.
- 1/2
- 1
- 0
- The limit does not exist
Correct answer: 0
The answer is 0. Rewrite (1 - cos x)/x by multiplying numerator and denominator by 1 + cos x, giving (sin² x)/x(1 + cos x). This equals (sin x)/x · (sin x)/(1 + cos x), which approaches 1 · 0/2 = 0. The value 1/2 is the known limit of (1 - cos x)/x², not divided by x.
- Evaluate lim x → -∞ (6 + 5x - 6x²)/(2x² + x - 1).
Correct answer: -3
The answer is -3. For a rational function where numerator and denominator share the same degree, the limit at -∞ equals the ratio of leading coefficients. The leading term of the numerator is -6x² and of the denominator is 2x², giving (-6)/2 = -3. The other choices ignore the leading-coefficient rule.
- Evaluate lim x → -∞ (√(9x² + 4))/(2x - 1).
Correct answer: -3/2
The answer is -3/2. For large magnitude x, √(9x² + 4) behaves like 3|x|. Since x → -∞, |x| = -x, so the numerator behaves like -3x while the denominator behaves like 2x. The ratio is (-3x)/(2x) = -3/2; the positive 3/2 ignores the sign change from the absolute value.
- Evaluate lim x → +∞ (2x² - 3x + 1)/(x - 5).
Correct answer: +∞
The answer is +∞. Because the numerator has degree 2 and the denominator has degree 1, the numerator grows without bound much faster than the denominator. Dividing the leading terms gives roughly 2x, which increases to +∞ as x grows. A finite limit of 2 or 0 would require equal or larger denominator degree.
- Consider f(x) = (x + 2)/(x² - x - 6). Determine lim x → -2 f(x).
- 1/5
- -1/5
- 0
- The limit does not exist
Correct answer: -1/5
The answer is -1/5. Factor the denominator as (x - 3)(x + 2) and cancel the common (x + 2) factor, leaving 1/(x - 3). Substituting x = -2 gives 1/(-5) = -1/5. The hole at x = -2 is removable, so the limit exists despite the original denominator being zero there.
- The function f(x) = (x + 1)/(x² - 1) has discontinuities at x = 1 and x = -1. Classify each discontinuity.
- Vertical asymptote at x = -1 and removable at x = 1
- Removable at both x = 1 and x = -1
- Removable at x = -1 and a vertical asymptote at x = 1
- Jump discontinuities at both points
Correct answer: Removable at x = -1 and a vertical asymptote at x = 1
The correct classification is removable at x = -1 and a vertical asymptote at x = 1. Factoring gives (x + 1)/((x - 1)(x + 1)); the (x + 1) factor cancels, creating a removable hole at x = -1. The remaining factor (x - 1) in the denominator cannot cancel, producing an infinite (vertical asymptote) discontinuity at x = 1. Rational functions do not produce jump discontinuities here.
- For the function f defined by f(x) = (x² - 16)/(x - 4) for x ≠ 4 and f(4) = c, what value of c makes f continuous at x = 4?
Correct answer: 8
The value is 8. To be continuous at x = 4, f(4) must equal lim x → 4 f(x). Factoring the numerator as (x - 4)(x + 4) and canceling leaves x + 4, whose limit at x = 4 is 8. Setting c = 8 removes the discontinuity; the other values do not match this limit.
- A function f is defined by f(x) = x² + 1 for x ≤ 2 and f(x) = ax - 1 for x > 2. For what value of a is f continuous at x = 2?
Correct answer: 3
The value is 3. Continuity at x = 2 requires the two pieces to agree there. The left piece gives 2² + 1 = 5, and the right piece gives a · 2 - 1 = 2a - 1. Setting 2a - 1 = 5 yields 2a = 6, so a = 3. The other values fail to match the left-hand value of 5.
- Determine lim x → 2 1/((x - 2)²).
- +∞
- -∞
- 0
- The limit does not exist because the one-sided limits differ
Correct answer: +∞
The answer is +∞. As x → 2 from either side, (x - 2)² is a small positive number because squaring removes any sign difference, so 1/((x-2)²) grows without bound. Both one-sided limits agree at +∞, so the limit is +∞ rather than -∞ or a case of disagreeing one-sided limits.
- Evaluate lim x → 0 (eˣ - 1)/x.
- The limit does not exist
- 0
- e
- 1
Correct answer: 1
The answer is 1. This is a standard limit reflecting that the slope of eˣ at x = 0 is 1; numerically, (eˣ - 1)/x approaches 1 as x nears 0 from both sides. The value e would be eˣ itself at x = 1, not this difference quotient, and 0 underestimates the behavior near the origin.
- A function g satisfies the inequality -x² ≤ g(x) ≤ x² for all x. Determine lim x → 0 g(x).
- 0
- 1
- -1
- The limit cannot be determined
Correct answer: 0
The answer is 0. By the Squeeze Theorem, since -x² and x² both approach 0 as x → 0, the function g trapped between them must also approach 0. Because both bounding functions share the same limit of 0, the limit of g is forced to that value, so it can indeed be determined.
- The function f is continuous on the closed interval [0, 4] with f(0) = -3 and f(4) = 5. According to the Intermediate Value Theorem, which statement must be true?
- The maximum value of f on the interval is 5
- f is increasing on the entire interval [0, 4]
- f(x) = 0 at exactly one value in the interval
- There is at least one value x between 0 and 4 where f(x) = 0
Correct answer: There is at least one value x between 0 and 4 where f(x) = 0
The correct statement is that there is at least one value x between 0 and 4 where f(x) = 0. Because f is continuous and 0 lies between f(0) = -3 and f(4) = 5, the Intermediate Value Theorem guarantees f takes the value 0 somewhere in the interval. The theorem does not require monotonic increase, a unique crossing, or that 5 is the maximum.
- Using the limit definition of the derivative, the difference quotient for f(x) = x² + 2x is set up as lim h → 0 ((x+h)² + 2(x+h) - (x² + 2x))/h. After expanding and simplifying, what is f'(x)?
Correct answer: 2x + 2
The answer is 2x + 2. Expanding the numerator gives x² + 2xh + h² + 2x + 2h - x² - 2x, which reduces to 2xh + h² + 2h. Dividing by h yields 2x + h + 2, and letting h → 0 leaves 2x + 2.
- What is the derivative of the constant function f(x) = 7?
Correct answer: 0
The answer is 0. The derivative of any constant function is zero because a constant graph is a horizontal line whose slope is zero everywhere, so the rate of change is 0.
- Find the derivative of f(x) = 3x⁷.
Correct answer: 21x⁶
The answer is 21x⁶. By the power rule, multiply the coefficient 3 by the exponent 7 to get 21, then reduce the exponent by one to get x⁶.
- A function f satisfies f(3) = 5, and its difference quotients near x = 3 approach the value 8 as the interval width shrinks to zero. What is f'(3)?
Correct answer: 8
The answer is 8. The derivative at a point is exactly the limit that the difference quotients approach as the interval width goes to zero, so f'(3) = 8. The value f(3) = 5 is the function value, not the slope.
- Find the derivative of f(x) = 10/x³ by first rewriting it with a negative exponent.
Correct answer: -30/x⁴
The answer is -30/x⁴. Rewrite f(x) as 10x⁻³; the power rule gives 10 · (-3) · x⁻⁴, which equals -30/x⁴.
- Find the derivative of f(x) = 5sin x - 3cos x.
- 5cos x - 3sin x
- 5cos x + 3sin x
- -5cos x + 3sin x
- 5sin x + 3cos x
Correct answer: 5cos x + 3sin x
The answer is 5cos x + 3sin x. The derivative of sin x is cos x, so 5sin x differentiates to 5cos x; the derivative of cos x is -sin x, so -3cos x differentiates to +3sin x.
- What is the derivative of f(x) = ln x?
Correct answer: 1/x
The answer is 1/x. The natural logarithm function ln x has derivative 1/x for x > 0, one of the basic derivative rules for common functions.
- Find the derivative of f(x) = eˣ sin x using the product rule.
- eˣ sin x
- eˣ cos x
- eˣ (cos x - sin x)
- eˣ (sin x + cos x)
Correct answer: eˣ (sin x + cos x)
The answer is eˣ (sin x + cos x). The product rule gives (eˣ)' sin x + eˣ (sin x)', which is eˣ sin x + eˣ cos x; factoring eˣ yields the result.
- Find the derivative of f(x) = (2x + 5)(x² - 1) using the product rule.
- 6x² + 10x - 2
- 4x + 0
- 2x² - 2
- 6x² - 2
Correct answer: 6x² + 10x - 2
The answer is 6x² + 10x - 2. The product rule gives 2(x² - 1) + (2x + 5)(2x), which is 2x² - 2 + 4x² + 10x, and combining like terms gives 6x² + 10x - 2.
- Find the derivative of f(x) = (3x + 1)/x using the quotient rule.
Correct answer: -1/x²
The answer is -1/x². The quotient rule gives (3 · x - (3x + 1) · 1)/x², which is (3x - 3x - 1)/x², equal to -1/x².
- What is the derivative of f(x) = tan x?
- sec x tan x
- -csc² x
- sec² x
- 1/(cos x)
Correct answer: sec² x
The answer is sec² x. The derivative of the tangent function is sec² x, a standard basic derivative rule for the trigonometric functions.
- Find the second derivative of f(x) = 2x³ - 4x + 9.
Correct answer: 12x
The answer is 12x. The first derivative is 6x² - 4, and differentiating again gives 12x, since the derivative of -4 is 0.
- Find the derivative of f(x) = 4√x by first writing it as a power of x.
Correct answer: 2/(√x)
The answer is 2/(√x). Write f(x) as 4x^(1/2); the power rule gives 4 · 1/2 · x^(-1/2), which is 2x^(-1/2), equal to 2/(√x).
- Evaluate the derivative of f(x) = x³ - 6x at x = 2.
Correct answer: 6
The answer is 6. The derivative is 3x² - 6, and substituting x = 2 gives 3 · 4 - 6, which is 12 - 6, equal to 6.
- A function f is differentiable at every real number. Which conclusion follows directly about f at each point in its domain?
- f is constant near that point
- f has a horizontal tangent at that point
- f is continuous at that point
- f has a corner at that point
Correct answer: f is continuous at that point
The answer is that f is continuous at that point. Differentiability at a point guarantees continuity there, because the limit defining the derivative can exist only where the function has no break, jump, or corner.
- Find the derivative of f(x) = (x² + 1)/(x + 2) using the quotient rule, then evaluate it at x = 0.
Correct answer: -1/4
The answer is -1/4. The quotient rule gives (2x(x + 2) - (x² + 1) · 1)/((x + 2)²); the numerator simplifies to 2x² + 4x - x² - 1, which is x² + 4x - 1. At x = 0 the numerator is -1 and the denominator (0 + 2)² is 4, so the value is -1/4.
- In the context of motion along a line, what does the derivative of a particle's position function with respect to time represent?
- The particle's velocity
- The particle's acceleration
- The total distance the particle has traveled
- The particle's average speed over the interval
Correct answer: The particle's velocity
Velocity is the answer because velocity is defined as the instantaneous rate of change of position with respect to time, which is exactly the derivative of the position function. Acceleration would be the second derivative, total distance involves an integral of speed, and average speed is a quotient over an interval rather than an instantaneous derivative.
- A cylinder has a fixed height of 10 centimeters, and its radius increases at 2 centimeters per second. Using V = π r² h, how fast is the volume increasing when the radius is 3 centimeters?
- 90π cubic centimeters per second
- 60π cubic centimeters per second
- 120π cubic centimeters per second
- 20π cubic centimeters per second
Correct answer: 120π cubic centimeters per second
The volume is increasing at 120π cubic centimeters per second. Differentiating V = π r² · 10 gives (dV)/dt = 20π r dr/dt; substituting r = 3 and dr/dt = 2 yields 20π · 3 · 2, which equals 120π. The other values come from dropping the factor of 2, mis-differentiating, or omitting the height.
- A function f satisfies f(9) = 3 and f'(9) = 1/6. Using the tangent-line approximation at x = 9, what is the estimate of f(9.6)?
Correct answer: 3.1
The estimate is 3.1. The linear approximation is L(x) = f(9) + f'(9)(x - 9), so L(9.6) = 3 + 1/6(0.6), which is 3 + 0.1, equal to 3.1. The other choices result from adding the wrong increment or using an incorrect slope.
- Evaluate lim x → 0 (eˣ - 1 - x)/x² using L'Hopital's Rule.
Correct answer: 1/2
The limit equals 1/2. The expression is the indeterminate form 0/0; one application of L'Hopital gives (eˣ - 1)/(2x), still 0/0, and a second application gives eˣ/2, which equals 1/2 at x = 0. The other values reflect stopping too early or arithmetic errors.
- A particle moves along a line with position s(t) = t³ - 3t² - 9t. On what interval is the particle slowing down?
- For all t > 3
- Only on the interval 0 < t < 1
- Only on the interval 1 < t < 3
- Only on the interval 0 < t < 3
Correct answer: Only on the interval 1 < t < 3
The particle is slowing down only on the interval 1 < t < 3. Velocity is 3t² - 6t - 9, which is zero at t = 3 (it is positive after); acceleration is 6t - 6, zero at t = 1. A particle slows down when velocity and acceleration have opposite signs, which holds on (1, 3). The other intervals mix matching signs.
- The area of an equilateral triangle is increasing at 8 square centimeters per second. The area in terms of side length is A = (√3)/4 s². How fast is the side length increasing when s = 4 centimeters?
- 4/(√3) centimeters per second
- 8/(√3) centimeters per second
- 2/(√3) centimeters per second
- √3 centimeters per second
Correct answer: 4/(√3) centimeters per second
The side length increases at 4/(√3) centimeters per second. Differentiating gives (dA)/dt = (√3)/2 s ds/dt; setting 8 = (√3)/2 · 4 · ds/dt gives 8 = 2√3 ds/dt, so ds/dt = 4/(√3). The other answers drop or double a factor.
- Why does using the tangent line at x = a to approximate a function near a generally produce a good estimate only for x values close to a?
- Because the function must be a polynomial for tangent lines to exist
- Because the derivative is undefined far from the point of tangency
- Because the tangent line always lies above the curve
- Because the tangent line and the curve share the same value and slope at a, but the curve bends away from the line as x moves farther from a
Correct answer: Because the tangent line and the curve share the same value and slope at a, but the curve bends away from the line as x moves farther from a
The correct reason is that the tangent line matches the curve's value and slope at a, but curvature causes the curve to bend away from the straight line as x moves farther from a, so the error grows with distance. The other statements are false: tangent lines exist for any differentiable function, do not always lie above the curve, and derivatives are not generally undefined away from a.
- Air is being pumped out of a spherical balloon so that its volume decreases at 36π cubic inches per second. Using (dV)/dt = 4π r² dr/dt, how fast is the radius changing when the radius is 3 inches?
- Decreasing at 4 inches per second
- Decreasing at 1 inch per second
- Increasing at 1 inch per second
- Decreasing at 3 inches per second
Correct answer: Decreasing at 1 inch per second
The radius is decreasing at 1 inch per second. Substituting into -36π = 4π · 9 · dr/dt gives -36π = 36π dr/dt, so dr/dt = -1, meaning the radius shrinks at 1 inch per second. The negative sign reflects the decreasing volume, ruling out the increasing option, and the other magnitudes come from arithmetic errors.
- Evaluate lim x → 0 (3ˣ - 1)/x using L'Hopital's Rule.
Correct answer: ln 3
The limit equals ln 3. The form is 0/0, and the derivative of 3ˣ is 3ˣ ln 3, while the derivative of x is 1; evaluating the ratio at x = 0 gives 3⁰ ln 3, which is ln 3. The other choices ignore the logarithm factor that comes from differentiating an exponential with base 3.
- A particle moves along a line so that its velocity is v(t) = 4 - 2t for t ≥ 0. What is the particle's speed at t = 4, and is the speed increasing or decreasing at that moment?
- Speed is 0 and increasing
- Speed is 4 and increasing
- Speed is 4 and decreasing
- Speed is -4 and increasing
Correct answer: Speed is 4 and increasing
The speed is 4 and increasing. At t = 4, velocity is 4 - 8 = -4, so speed is the absolute value 4; acceleration is -2, and since velocity and acceleration are both negative they share a sign, meaning the particle is speeding up. Speed cannot be negative, ruling out that choice, and it is not zero or decreasing here.
- Using a local linearization of f(x) = (1 + x)^(1/3) centered at x = 0, estimate (1.06)^(1/3).
Correct answer: 1.02
The estimate is 1.02. The derivative is 1/3(1 + x)^(-2/3), which equals 1/3 at x = 0, so L(x) = 1 + 1/3x; substituting x = 0.06 gives 1 + 0.02, equal to 1.02. The other choices come from using the wrong slope or simply restating the input.
- A streetlight is mounted at the top of a pole, and a person walks away from the base of the pole at a constant speed. In a related rates analysis of the length of the person's shadow, which quantities are changing with respect to time?
- The person's distance from the pole and the length of the shadow
- Only the angle of elevation of the sun, which is fixed
- The height of the person, which decreases over time
- Only the height of the pole
Correct answer: The person's distance from the pole and the length of the shadow
The changing quantities are the person's distance from the pole and the length of the shadow. As the person walks, both the distance to the pole and the shadow length vary with time, and they are linked by similar triangles, allowing differentiation with respect to time. The pole height and the person's height are fixed constants, so they are not changing.
- Evaluate lim x → 0⁺ (x ln x) by first rewriting it for L'Hopital's Rule.
Correct answer: 0
The limit equals 0. The product is the indeterminate form 0 · (-∞), so it is rewritten as (ln x)/(1/x), an ∞/∞ form; applying L'Hopital gives (1/x)/(-1/x²), which simplifies to -x and approaches 0. The other values ignore the rewriting step or mis-simplify the resulting quotient.
- A particle moving along a line has acceleration that is always positive while its velocity is negative throughout an interval. What is happening to the particle's speed on that interval?
- The speed is constant
- The speed is decreasing
- The speed is increasing
- The speed is zero
Correct answer: The speed is decreasing
The speed is decreasing. When acceleration and velocity have opposite signs, the particle is slowing down, so a positive acceleration paired with a negative velocity means the speed is decreasing toward zero. Matching signs would mean speeding up, and neither constant nor zero speed follows from opposite-signed acceleration and velocity.
- The temperature T of an object, in degrees, is modeled so that at a certain instant (dT)/dt = -4 degrees per minute. What is the correct interpretation of this value in context?
- The object's temperature increases by 4 degrees total
- The object will reach 4 degrees in one minute
- The object's temperature is decreasing at a rate of 4 degrees per minute at that instant
- The object's temperature is 4 degrees at that instant
Correct answer: The object's temperature is decreasing at a rate of 4 degrees per minute at that instant
The correct interpretation is that the temperature is decreasing at a rate of 4 degrees per minute at that instant. A derivative (dT)/dt gives an instantaneous rate of change, and the negative sign indicates the temperature is falling. The other statements confuse the rate with the temperature value itself or with a total change rather than an instantaneous rate.
- Two cars leave the same intersection at the same time, one heading north and one heading east. In setting up the related rate for how fast the distance between them changes, which equation should be differentiated with respect to time?
- x + y = z
- x² + y² = z², where x and y are the distances traveled and z is the distance between the cars
- xy = z
- x² + y² = z
Correct answer: x² + y² = z², where x and y are the distances traveled and z is the distance between the cars
The correct equation is x² + y² = z², the Pythagorean relationship, because the two perpendicular paths and the straight-line distance between the cars form a right triangle. Differentiating it with respect to time relates all three rates. The other equations do not correctly model the right-triangle geometry of perpendicular travel.
- Evaluate lim x → 0 (cos x - 1)/(x² + x) using L'Hopital's Rule.
Correct answer: 0
The limit equals 0. The form is 0/0, and one application of L'Hopital gives (-sin x)/(2x + 1); at x = 0 the numerator is 0 and the denominator is 1, so the value is 0. The other choices come from incorrectly treating the denominator as x² alone or from a sign error.
- The position of a particle is given by s(t) = t² - 6t + 8 for t ≥ 0, where s is in meters and t is in seconds. What is the displacement of the particle from t = 0 to t = 5?
- -5 meters
- 3 meters
- 5 meters
- 13 meters
Correct answer: -5 meters
The displacement is -5 meters. Displacement is the change in position, s(5) - s(0); s(5) = 25 - 30 + 8 = 3, and s(0) = 8, so the difference is 3 - 8 = -5 meters. The other values reverse the subtraction, use only one position value, or sum them.
- Using a tangent-line approximation, the value of a differentiable function that is concave down near the point of tangency is estimated. How does this linear estimate compare with the true function value for inputs near the point of tangency?
- The estimate is less than the true value (an underestimate)
- The estimate is greater than the true value (an overestimate)
- The comparison cannot be determined from concavity
- The estimate exactly equals the true value
Correct answer: The estimate is greater than the true value (an overestimate)
The estimate is an overestimate because the function is concave down. When a curve is concave down, it lies below its tangent line near the point of tangency, so the tangent-line value sits above the actual function value, making the linear approximation too large. Concave-up curves would give underestimates, and concavity does determine the direction of the error.
- Rolle's Theorem applies to f(x) = x² - 6x + 8 on the closed interval [2, 4]. Find the value of c in the open interval that the theorem guarantees.
Correct answer: c = 3
The guaranteed value is c = 3. Because f(2) and f(4) both equal 0, Rolle's Theorem guarantees an interior point where f' equals zero. Setting f'(x) = 2x - 6 = 0 gives x = 3, which lies in the open interval (2, 4). The endpoints 2 and 4 are excluded, and 3.5 does not make the derivative zero.
- For f(x) = x³ on the closed interval [0, 3], find the value of c guaranteed by the Mean Value Theorem.
Correct answer: c = √3
The guaranteed value is c = √3. The average rate of change is (f(3) - f(0))/(3 - 0) = (27 - 0)/3 = 9. Setting f'(x) = 3x² = 9 gives x² = 3, so x = √3, which lies in the interval. The midpoint 1.5 and the endpoint 3 do not satisfy the equation.
- A function f is differentiable on the entire real line and f'(x) is negative for all x < 5 and positive for all x > 5. What can be concluded about x = 5?
- f has a relative minimum at x = 5
- f has a relative maximum at x = 5
- f has an inflection point at x = 5
- f has neither a maximum nor a minimum at x = 5
Correct answer: f has a relative minimum at x = 5
The answer is that f has a relative minimum at x = 5. By the First Derivative Test, when f' changes from negative to positive as x increases through a critical point, the function changes from decreasing to increasing, producing a relative minimum. A sign change from positive to negative would give a maximum, and an inflection point concerns concavity, not the sign of the first derivative.
- On what interval is the function f(x) = x³ - 12x increasing?
- x < -2 and x > 2
- The interval (-2, 2)
- x > 0 only
- The entire real line
Correct answer: x < -2 and x > 2
The function is increasing for x < -2 and for x > 2. The derivative f'(x) = 3x² - 12 = 3(x - 2)(x + 2). This is positive when x < -2 or x > 2, where the function increases. Between -2 and 2 the derivative is negative, so the function decreases there.
- The graph of f has f''(x) = 12x² - 24x. On what interval is the graph of f concave up?
- x < 0 and x > 2
- The interval (0, 2)
- x > 1 only
- The entire real line
Correct answer: x < 0 and x > 2
The graph is concave up for x < 0 and for x > 2. Factoring gives f''(x) = 12x(x - 2), which is positive when x < 0 or x > 2. Concavity is upward where the second derivative is positive. On the interval (0, 2) the second derivative is negative, indicating concave down.
- A rectangle is inscribed with its base on the x-axis under the parabola y = 9 - x², with vertices symmetric about the y-axis. What value of x maximizes the area of the rectangle?
Correct answer: x = √3
The maximizing value is x = √3. The rectangle has width 2x and height 9 - x², so area A = 2x(9 - x²) = 18x - 2x³. Setting A' = 18 - 6x² = 0 gives x² = 3, so x = √3. The value 3 gives zero height, and the other choices do not satisfy the optimization equation.
- Find the absolute maximum value of f(x) = 2x³ - 9x² + 12x on the closed interval [0, 3].
Correct answer: 9
The absolute maximum value is 9, attained at x = 3. Critical points come from f'(x) = 6x² - 18x + 12 = 6(x - 1)(x - 2), giving x = 1 and x = 2. Evaluating f at 0, 1, 2, and 3 gives 0, 5, 4, and 9 respectively. The largest of these candidate values is 9 at the right endpoint.
- A cylindrical can with a closed top and bottom must have a volume of 16π cubic centimeters. What radius minimizes the total surface area?
- r = 2
- r = 4
- r = 16
- r = ³√(16)
Correct answer: r = 2
The minimizing radius is r = 2. With volume π r² h = 16π, the height is h = 16/r². Surface area S = 2π r² + 2π r h = 2π r² + (32π)/r. Setting S' = 4π r - (32π)/r² = 0 gives r³ = 8, so r = 2. The other values do not satisfy the minimization condition.
- Using the Second Derivative Test, classify the critical point of f(x) = x³ - 12x at x = 2.
- Relative minimum because f''(2) is positive
- Relative maximum because f''(2) is negative
- Inflection point because f''(2) = 0
- Neither, because the test is inconclusive
Correct answer: Relative minimum because f''(2) is positive
The point at x = 2 is a relative minimum because f''(2) is positive. The first derivative f'(x) = 3x² - 12 is zero at x = 2. The second derivative f''(x) = 6x evaluated at 2 gives 12, which is positive, so the graph is concave up there, indicating a relative minimum.
- The function f(x) = x⁴ has f'(0) = 0. Why does the First Derivative Test, rather than the Second Derivative Test, more readily classify x = 0?
- f''(0) = 0, making the Second Derivative Test inconclusive, but f' changes from negative to positive
- f''(0) is negative, so the Second Derivative Test gives a maximum
- f' does not change sign, so neither test applies
- f is not differentiable at x = 0
Correct answer: f''(0) = 0, making the Second Derivative Test inconclusive, but f' changes from negative to positive
The reason is that f''(0) = 0, making the Second Derivative Test inconclusive, while the First Derivative Test still works because f' changes from negative to positive. Here f''(x) = 12x² is zero at the origin, giving no information. However f'(x) = 4x³ is negative for x < 0 and positive for x > 0, so x = 0 is a relative minimum.
- Find the x-coordinate of the inflection point of f(x) = x³ - 6x² + 9x + 2.
Correct answer: x = 2
The inflection point occurs at x = 2. The second derivative f''(x) = 6x - 12 = 0 when x = 2. Checking shows f'' is negative for x < 2 and positive for x > 2, so concavity changes there. The values 1 and 3 are critical points of the first derivative, not the second.
- A point moves along the curve y = √x. At what x-value is the function increasing most rapidly per unit increase in x compared to elsewhere on the domain?
- As x → 0⁺, where the derivative grows without bound
- At x = 1, where the derivative equals 1/2
- At x = 4, where the derivative equals 1/4
- As x → ∞, where the derivative is largest
Correct answer: As x → 0⁺, where the derivative grows without bound
The function increases most rapidly as x → 0⁺, where the derivative grows without bound. The derivative of √x is 1/(2√x), which becomes arbitrarily large as x → 0⁺. At x = 1 it is 1/2 and at x = 4 it is 1/4, and the derivative decreases toward 0 as x grows, so larger x gives a smaller rate.
- A continuous function f is defined on the closed interval [-1, 4]. It has critical points at x = 1 and x = 3. Which set of points must be checked to find the absolute extrema?
- x = -1, 1, 3, and 4
- x = 1 and 3 only
- x = -1 and 4 only
- x = 0, 1, 3, and 4
Correct answer: x = -1, 1, 3, and 4
The points to check are x = -1, 1, 3, and 4. The Closed Interval Method requires evaluating f at every interior critical point and at both endpoints, then comparing values. The critical points are 1 and 3, and the endpoints are -1 and 4. Omitting endpoints or interior critical points could miss the true absolute maximum or minimum.
- The Candidates Test gives the absolute minimum of f(x) = x + 4/x on the closed interval [1, 6]. What is that minimum value?
- 4, attained at x = 2
- 5, attained at x = 1
- 71/3, attained at x = 6
- 0, attained at x = 2
Correct answer: 4, attained at x = 2
The absolute minimum value is 4, attained at x = 2. The derivative f'(x) = 1 - 4/x² is zero when x² = 4, so x = 2 lies in the interval. Evaluating f at 1, 2, and 6 gives 5, 4, and 62/3. The smallest of these candidate values is 4 at x = 2.
- If f'(x) = (x - 1)² (x - 4), at which x-value does f have a relative minimum?
- x = 4
- x = 1
- x = 0
- f has no relative minimum
Correct answer: x = 4
The relative minimum occurs at x = 4. The factor (x - 1)² is always nonnegative, so the sign of f' is governed by (x - 4): negative for x < 4 and positive for x > 4, a change from negative to positive giving a minimum. At x = 1 the squared factor does not change sign, so f' does not change sign there and no extremum occurs.
- A 10-foot wire is cut into two pieces; one piece is bent into a square and the other into a circle. To minimize the total enclosed area, what length should be used for the square's wire? (Approximately, using that the square's piece is 40/(4 + π) feet.)
- About 5.6 feet, found by setting the derivative of total area to zero
- Exactly 5 feet, splitting the wire evenly
- 0 feet, using all wire for the circle
- All 10 feet for the square
Correct answer: About 5.6 feet, found by setting the derivative of total area to zero
The square's piece should be about 5.6 feet, found by setting the derivative of total area to zero. If x feet make the square, its area is (x/4)², and the remaining 10 - x make a circle of area found from circumference 10 - x. Differentiating the total area and setting it to zero yields x = 40/(4 + π), which is approximately 5.6 feet. Splitting evenly or using all wire for one shape does not satisfy the optimization condition.
- A function f satisfies f'(x) > 0 and f''(x) < 0 on an interval. Which description fits the graph there?
- Increasing and concave down
- Increasing and concave up
- Decreasing and concave down
- Decreasing and concave up
Correct answer: Increasing and concave down
The graph is increasing and concave down on that interval. A positive first derivative means the function values rise as x increases, so the graph is increasing. A negative second derivative means the slope is decreasing, so the graph bends downward and is concave down. The remaining combinations contradict at least one of the given sign conditions.
- Find the dimensions x by x by h of an open-top box with a square base and volume 32 cubic units that minimizes the surface area. What base side x minimizes the material used?
Correct answer: x = 4
The minimizing base side is x = 4. With volume x² h = 32, the height is h = 32/x². The open-top surface area is S = x² + 4xh = x² + 128/x. Setting S' = 2x - 128/x² = 0 gives x³ = 64, so x = 4. The other values do not satisfy the minimization equation.
- The graph of f' is a straight line that crosses the x-axis at x = 3, is negative before 3, and positive after. What does this tell you about f at x = 3?
- f has a relative minimum at x = 3
- f has a relative maximum at x = 3
- f has an inflection point at x = 3
- f is not defined at x = 3
Correct answer: f has a relative minimum at x = 3
The graph of f has a relative minimum at x = 3. Reading the graph of f', the first derivative changes from negative to positive at x = 3, so f changes from decreasing to increasing. By the First Derivative Test this sign change produces a relative minimum. An inflection point would be read from where f' has a local extremum, not where it crosses zero.
- For f(x) = sin x on the closed interval [0, π], find the value of c guaranteed by the Mean Value Theorem.
Correct answer: c = π/2
The guaranteed value is c = π/2. The average rate of change is (sin π - sin 0)/(π - 0) = (0 - 0)/π = 0. Setting f'(x) = cos x = 0 on the interval gives x = π/2. The endpoints give nonzero cosine values, so they do not qualify.
- A function has f''(x) = (x - 1)(x - 5). How many inflection points does the graph of f have?
- Two, at x = 1 and x = 5
- One, at x = 3
- None, because the second derivative is a product
- Three, at x = 1, 3, and 5
Correct answer: Two, at x = 1 and x = 5
The graph has two inflection points, at x = 1 and x = 5. The second derivative (x - 1)(x - 5) changes sign at each of its simple roots: positive before 1, negative between 1 and 5, and positive after 5. Each sign change marks a change in concavity and thus an inflection point. The value x = 3 is where the second derivative reaches a minimum, not where concavity changes.
- Determine the interval on which f(x) = x e⁻ˣ is increasing.
- x < 1
- x > 1
- The entire real line
- x < 0
Correct answer: x < 1
The function is increasing for x < 1. Using the product rule, f'(x) = e⁻ˣ(1 - x). Since e⁻ˣ is always positive, the sign of f' matches the sign of (1 - x), which is positive when x < 1. For x > 1 the derivative is negative, so the function decreases there.
- Evaluate the indefinite integral ∫ (4x³ - 6x + 5) dx.
- 4x⁴ - 3x² + 5x + C
- x⁴ - 3x² + 5x + C
- 12x² - 6 + C
- x⁴ - 6x² + 5x + C
Correct answer: x⁴ - 3x² + 5x + C
The antiderivative is x⁴ - 3x² + 5x + C. Apply the power rule term by term: the integral of 4x³ is x⁴, the integral of -6x is -3x², the integral of 5 is 5x, and you must add the constant C for an indefinite integral.
- Evaluate the definite integral ∫ from 1 to 2 of 3x² dx.
Correct answer: 7
The value is 7. An antiderivative of 3x² is x³, so by the Fundamental Theorem of Calculus evaluate x³ at 2 and at 1: 2³ = 8 and 1³ = 1, and 8 - 1 = 7.
- A trapezoidal sum is used to approximate the area under a curve that is concave up on the entire interval. Compared with the exact integral, this approximation will be:
- Exactly equal, because trapezoids account for the slope of the curve
- Undetermined without knowing the sign of the function
- An underestimate, because the trapezoids lie below a concave-up curve
- An overestimate, because the trapezoids lie above a concave-up curve
Correct answer: An overestimate, because the trapezoids lie above a concave-up curve
The trapezoidal sum is an overestimate. When a function is concave up, the straight chord connecting two points on the curve lies above the curve between those points, so each trapezoid encloses slightly more area than the true region, producing an overestimate.
- Use u-substitution to evaluate the indefinite integral ∫ 3x² √(x³ + 2) dx.
- 2/9(x³ + 2)^(3/2) + C
- (x³ + 2)^(3/2) + C
- 2/3(x³ + 2)^(3/2) + C
- 3/2(x³ + 2)^(3/2) + C
Correct answer: 2/3(x³ + 2)^(3/2) + C
The result is 2/3(x³ + 2)^(3/2) + C. Let u = x³ + 2, so du = 3x² dx; the integral becomes ∫ √u du, which is 2/3 u^(3/2), then substitute u back.
- Evaluate the definite integral ∫ from 1 to e of 1/x dx.
Correct answer: 1
The value is 1. An antiderivative of 1/x is ln|x|, so evaluate ln x from 1 to e: ln e = 1 and ln 1 = 0, and 1 - 0 = 1.
- Find the derivative with respect to x of the accumulation function F(x) = ∫ from 2 to x of cos(t²) dt.
- -sin(x²) · 2x
- cos(x²)
- cos(x²) · 2x
- sin(x²)
Correct answer: cos(x²)
The derivative is cos(x²). By the Fundamental Theorem of Calculus, the derivative of ∫ from a to x of f(t) dt is simply f(x), so you substitute the upper limit x into the integrand cos(t²).
- The rate at which water flows into a tank is given by R(t) liters per minute. Which expression represents the total amount of water that flows into the tank from t = 0 to t = 10 minutes?
- R(10) - R(0)
- R'(10)
- ∫ from 0 to 10 of R(t) dt
- R(10)/10
Correct answer: ∫ from 0 to 10 of R(t) dt
The total water is ∫ from 0 to 10 of R(t) dt. Integrating a rate of change over an interval gives the net accumulation of the quantity, so integrating the flow rate over the ten minutes yields the total volume that flowed in.
- What is the average value of the function f(x) = 4x on the closed interval [0, 3]?
Correct answer: 6
The average value is 6. Use the formula 1/(b - a)∫ from a to b of f(x) dx: the integral ∫ from 0 to 3 of 4x dx = 2x² evaluated gives 18, and 18/(3 - 0) = 6.
- Use integration by parts to evaluate the indefinite integral ∫ x eˣ dx.
- x²/2 eˣ + C
- eˣ - x eˣ + C
- x eˣ - eˣ + C
- x eˣ + eˣ + C
Correct answer: x eˣ - eˣ + C
The result is x eˣ - eˣ + C. With integration by parts let u = x and dv = eˣ dx, so the integral equals x eˣ - ∫ eˣ dx, which gives x eˣ - eˣ.
- A right Riemann sum with three subintervals of equal width approximates ∫ from 0 to 6 of f(x) dx using the values f(2)=5, f(4)=8, and f(6)=11. What is the approximation?
Correct answer: 48
The approximation is 48. The width of each subinterval is 2, and a right sum uses the right endpoints f(2), f(4), and f(6), giving 2(5+8+11)=2· 24=48.
- Evaluate the definite integral ∫ from 0 to π of sin x dx.
Correct answer: 2
The value is 2. An antiderivative of sin x is -cos x, so evaluate -cos x from 0 to π: -cos π = 1 and -cos 0 = -1, and 1-(-1)=2.
- Find the indefinite integral ∫ sec² x dx.
- sec x tan x + C
- tan x + C
- -cot x + C
- sec x + C
Correct answer: tan x + C
The antiderivative is tan x + C. This is a standard antiderivative because the derivative of tan x is sec² x, so reversing that differentiation gives tan x + C.
- If ∫ from 1 to 5 of f(x) dx = 12 and ∫ from 1 to 3 of f(x) dx = 7, what is the value of ∫ from 3 to 5 of f(x) dx?
Correct answer: 5
The value is 5. By the additivity property of definite integrals, ∫ from 1 to 3 of f(x) dx + ∫ from 3 to 5 of f(x) dx = ∫ from 1 to 5 of f(x) dx, so ∫ from 3 to 5 of f(x) dx = 12 - 7 = 5.
- Evaluate the indefinite integral ∫ 1/(x² + 1) dx.
- arcsin x + C
- arctan x + C
- ln(x² + 1) + C
- -1/(x² + 1) + C
Correct answer: arctan x + C
The antiderivative is arctan x + C. This is a standard inverse-trigonometric antiderivative, because the derivative of arctan x is exactly 1/(x² + 1).
- A particle moves along a line with velocity v(t) = 6t. What is the total displacement of the particle from t = 0 to t = 2?
Correct answer: 12
The displacement is 12. Displacement equals ∫ from 0 to 2 of 6t dt = 3t²|₀² = 3· 4 - 0 = 12.
- Use u-substitution to evaluate the indefinite integral ∫ cos(5x) dx.
- 1/5sin(5x) + C
- -1/5sin(5x) + C
- 5sin(5x) + C
- sin(5x) + C
Correct answer: 1/5sin(5x) + C
The result is 1/5sin(5x) + C. Let u = 5x so du = 5 dx; the integral becomes 1/5∫ cos u du = 1/5sin u, then substitute u = 5x.
- The graph of a continuous function f lies entirely below the x-axis on the interval from a to b. What can be concluded about the definite integral ∫ from a to b of f(x) dx?
- It is negative, because the signed area below the axis counts as negative
- It is zero, because the regions cancel
- It cannot be determined without the formula for f
- It is positive, because area is always nonnegative
Correct answer: It is negative, because the signed area below the axis counts as negative
The integral is negative. A definite integral measures signed area, so any portion of the region lying below the x-axis contributes a negative amount; if the entire graph is below the axis, the total signed area is negative.
- Evaluate the definite integral ∫ from 0 to 4 of (2x + 3) dx.
Correct answer: 28
The value is 28. An antiderivative of 2x + 3 is x² + 3x, so evaluate from 0 to 4: at x = 4 you get 16 + 12 = 28, and at x = 0 you get 0, giving 28 - 0 = 28.
- Let g(x) = ∫ from 0 to x² of f(t) dt. Using the chain rule with the Fundamental Theorem of Calculus, what is g'(x)?
- f(x²)· 2x
- f(x)· 2x
- 2x· f(t)
- f(x²)
Correct answer: f(x²)· 2x
The derivative is f(x²)· 2x. Because the upper limit is x² rather than x, the Fundamental Theorem combined with the chain rule gives f evaluated at the upper limit, f(x²), multiplied by the derivative of x², which is 2x.
- Find the indefinite integral ∫ e⁻ˣ dx.
- -e⁻ˣ + C
- e⁻ˣ + C
- x e⁻ˣ + C
- -1/x e⁻ˣ + C
Correct answer: -e⁻ˣ + C
The antiderivative is -e⁻ˣ + C. Differentiating -e⁻ˣ by the chain rule returns e⁻ˣ, confirming it is the correct antiderivative of e⁻ˣ.
- A left Riemann sum and a right Riemann sum are computed for a continuous, increasing function on the same interval with the same equal subintervals. Which statement is true?
- The left sum overestimates and the right sum underestimates the integral
- Both sums give exactly the integral
- Both sums overestimate the integral
- The left sum underestimates and the right sum overestimates the integral
Correct answer: The left sum underestimates and the right sum overestimates the integral
For an increasing function the left sum underestimates and the right sum overestimates. Each left endpoint gives the smallest function value on its subinterval producing rectangles below the curve, while each right endpoint gives the largest value producing rectangles above the curve.
- Evaluate the indefinite integral ∫ x⁻² dx, that is ∫ 1/x² dx.
- ln(x²) + C
- -2/x³ + C
- -1/x + C
- 1/x + C
Correct answer: -1/x + C
The antiderivative is -1/x + C. Rewrite 1/x² as x⁻²; the power rule adds one to the exponent giving x⁻¹/(-1), which equals -1/x.
- If g(x) = ∫ from 0 to x of f(t) dt and f is positive on the interval shown, what can be said about g on that interval?
- g must equal zero throughout the interval
- g is constant, because f does not change sign
- g is increasing, because its derivative f is positive
- g is decreasing, because the integral subtracts area
Correct answer: g is increasing, because its derivative f is positive
The function g is increasing. By the Fundamental Theorem of Calculus, g'(x) = f(x); since f is positive on the interval, g' is positive there, which means g is increasing.
- Use u-substitution to evaluate the definite integral ∫ from 0 to 1 of 2x(x² + 1)³ dx.
Correct answer: 15/4
The value is 15/4. Let u = x² + 1 so du = 2x dx; the limits become u = 1 and u = 2, and ∫ u³ du = u⁴/4, giving (16 - 1)/4 = 15/4.
- The acceleration of an object is a(t) = 4 and its velocity at t = 0 is v(0) = 3. Using accumulation of the acceleration, what is the velocity v(t)?
Correct answer: 4t + 3
The velocity is 4t + 3. Velocity is the antiderivative of acceleration, so integrating the constant 4 gives 4t plus a constant; using the initial condition v(0) = 3 fixes the constant at 3, giving 4t + 3.
- Evaluate the definite integral ∫ from 0 to 2 of eˣ dx.
Correct answer: e² - 1
The value is e² - 1. An antiderivative of eˣ is eˣ itself, so evaluate from 0 to 2: e² - e⁰, and since e⁰ = 1, the result is e² - 1.
- Which property correctly describes the definite integral ∫ from a to a of f(x) dx, where the lower and upper limits are equal?
- It equals 0, because the interval has zero width
- It equals f(a), because only one point contributes
- It is undefined, because the limits coincide
- It equals the area under the entire curve of f
Correct answer: It equals 0, because the interval has zero width
The integral equals 0. When the upper and lower limits of integration are identical, the region has no width and therefore encloses zero area, so the definite integral is 0 regardless of the integrand.
- What is the average value of the function f(x) = x² on the interval from x = 0 to x = 3?
Correct answer: 3
The average value is 3. The average value of a function on [a,b] is 1/(b-a)∫ from a to b of f(x) dx. Here that is 1/3∫ from 0 to 3 of x² dx = 1/3·27/3 = 1/3· 9 = 3.
- An object moves along a line with velocity v(t) = 6t feet per second. What is the total displacement of the object from t = 0 to t = 4 seconds?
- 24 feet
- 48 feet
- 12 feet
- 96 feet
Correct answer: 48 feet
The displacement is 48 feet. Displacement equals ∫ from 0 to 4 of 6t dt. The antiderivative is 3t², evaluated from 0 to 4 gives 3· 16 - 0 = 48 feet.
- A particle has velocity v(t) = t - 3 meters per second for t between 0 and 5 seconds. What is the total distance traveled by the particle over this interval?
- 2.5 meters
- 5 meters
- 6.5 meters
- 12.5 meters
Correct answer: 6.5 meters
The total distance traveled is 6.5 meters. Total distance uses ∫ from 0 to 5 of |v(t)| dt. The velocity is negative from 0 to 3 and positive from 3 to 5. The integral ∫ from 0 to 3 of (3 - t) dt gives an area of 4.5, and ∫ from 3 to 5 of (t - 3) dt gives an area of 2. Adding the magnitudes 4.5 and 2 yields 6.5 meters.
- The rate at which water flows into a tank is given by R(t) = 4t gallons per minute. How many gallons enter the tank during the first 5 minutes?
- 20 gallons
- 100 gallons
- 25 gallons
- 50 gallons
Correct answer: 50 gallons
The amount that enters is 50 gallons. The total accumulation equals ∫ from 0 to 5 of 4t dt. The antiderivative is 2t², evaluated from 0 to 5 gives 2· 25 - 0 = 50 gallons.
- If a tank initially holds 10 liters and water flows in at a rate of R(t) = 3 liters per minute, which expression gives the amount of water in the tank at time t = 6 minutes?
- 10 + ∫ from 0 to 6 of 3 dt
- ∫ from 0 to 6 of 3 dt
- 10 - ∫ from 0 to 6 of 3 dt
- 10 · ∫ from 0 to 6 of 3 dt
Correct answer: 10 + ∫ from 0 to 6 of 3 dt
The correct expression is 10 + ∫ from 0 to 6 of 3 dt. The accumulation function adds the initial amount to the net change, where net change is the integral of the inflow rate. The starting 10 liters is the initial condition and the integral accounts for what is added over the six minutes.
- Find the area of the region in the first quadrant bounded by y = 4 - x² and the x-axis.
Correct answer: 16/3
The area is 16/3. In the first quadrant the region runs from x = 0 to x = 2, where y = 4 - x² meets the x-axis. The area is ∫ from 0 to 2 of (4 - x²) dx. The antiderivative is 4x - x³/3; evaluating from 0 to 2 gives 8 - 8/3 = 16/3.
- The region bounded by y = eˣ, the x-axis, x = 0, and x = 1 is revolved about the x-axis. Which integral gives the volume using the disk method?
- π ∫ from 0 to 1 of eˣ dx
- 2π ∫ from 0 to 1 of x eˣ dx
- π ∫ from 0 to 1 of (eˣ)² dx
- ∫ from 0 to 1 of (eˣ)² dx
Correct answer: π ∫ from 0 to 1 of (eˣ)² dx
The correct integral is π ∫ from 0 to 1 of (eˣ)² dx. The disk method uses π times the integral of the radius squared, and here the radius of each disk is the function value eˣ. Squaring it and multiplying by π gives the volume integral.
- A solid has a base that is the region bounded by y = x and the x-axis from x = 0 to x = 4, and every cross section perpendicular to the x-axis is a square. Which integral gives the volume?
- ∫ from 0 to 4 of x dx
- π ∫ from 0 to 4 of x² dx
- ∫ from 0 to 4 of (4 - x)² dx
- ∫ from 0 to 4 of x² dx
Correct answer: ∫ from 0 to 4 of x² dx
The correct integral is ∫ from 0 to 4 of x² dx. For square cross sections the area of each slice is the side length squared. The side length equals the height of the region, which is x, so each cross-sectional area is x² and the volume is the integral of that area.
- The region bounded by y = x² and y = x is revolved about the x-axis. Which integral gives the volume using the washer method?
- π ∫ from 0 to 1 of (x² - x⁴) dx
- π ∫ from 0 to 1 of (x - x²) dx
- π ∫ from 0 to 1 of (x⁴ - x²) dx
- 2π ∫ from 0 to 1 of (x - x²) dx
Correct answer: π ∫ from 0 to 1 of (x² - x⁴) dx
The correct integral is π ∫ from 0 to 1 of (x² - x⁴) dx. The curves intersect at x = 0 and x = 1, where y = x is the outer radius and y = x² is the inner radius. The washer area is π(outer² - inner²) = π(x² - x⁴).
- What is the average value of f(x) = cos x on the interval from x = 0 to x = π/2?
Correct answer: 2/π
The average value is 2/π. The average value equals 1/(π/2 - 0)∫ from 0 to π/2 of cos x dx. The integral of cos x is sin x, which evaluates to 1 - 0 = 1, and multiplying by 2/π gives 2/π.
- A particle moving along the x-axis has acceleration a(t) = 6t and at t = 0 its velocity is 2. What is its velocity at t = 2?
Correct answer: 14
The velocity at t = 2 is 14. Velocity is the antiderivative of acceleration plus the initial velocity: ∫ 6t dt = 3t², so v(t) = 3t² + 2 using v(0) = 2. At t = 2 this gives 3· 4 + 2 = 14.
- Find the area of the region enclosed by the curves y = x³ and y = x.
Correct answer: 1/2
The area is 1/2. The curves intersect at x = -1, 0, and 1, and by symmetry the total enclosed area is twice the area from 0 to 1. On that interval x is above x³, so ∫ from 0 to 1 of (x - x³) dx = 1/2 - 1/4 = 1/4; doubling for both regions gives 1/2.
- The region bounded by y = 2x, the x-axis, and x = 3 is revolved about the x-axis. What is the volume of the resulting solid?
Correct answer: 36π
The volume is 36π. Using the disk method, the volume is π ∫ from 0 to 3 of (2x)² dx = π ∫ from 0 to 3 of 4x² dx. The antiderivative is (4x³)/3, evaluated from 0 to 3 gives (4· 27)/3 = 36, so the volume is 36π.
- If F(x) = ∫ from 0 to x of √(t² + 1) dt, what does F(x) represent for x > 0?
- The slope of the tangent line to √(t² + 1) at x
- The accumulated area under the curve y = √(t² + 1) from 0 to x
- The average value of √(t² + 1) on the interval
- The volume of revolution of the curve about the t-axis
Correct answer: The accumulated area under the curve y = √(t² + 1) from 0 to x
F(x) represents the accumulated area under y = √(t² + 1) from 0 to x. An accumulation function defined as a definite integral with a variable upper limit gives the net signed area between the lower limit and the upper limit. Since the integrand is positive, this is the actual area collected from 0 out to x.
- A region is bounded by y = 1/x, the x-axis, x = 1, and x = e. What is the area of this region?
Correct answer: 1
The area is 1. The area equals ∫ from 1 to e of 1/x dx. The antiderivative of 1/x is ln x, evaluated from 1 to e gives ln e - ln 1 = 1 - 0 = 1.
- The base of a solid is the region bounded by y = 2 - x and the coordinate axes in the first quadrant. Each cross section perpendicular to the x-axis is a semicircle with diameter in the base. Which integral gives the volume?
- ∫ from 0 to 2 of π/2(2 - x)² dx
- ∫ from 0 to 2 of π(2 - x)² dx
- ∫ from 0 to 2 of (2 - x)² dx
- ∫ from 0 to 2 of π/8(2 - x)² dx
Correct answer: ∫ from 0 to 2 of π/8(2 - x)² dx
The correct integral is ∫ from 0 to 2 of π/8(2 - x)² dx. The diameter of each semicircle is the height (2 - x), so the radius is (2 - x)/2. A semicircle area is 1/2π r² = 1/2π((2 - x)/2)², simplifying to π/8(2 - x)².
