Drill 1 · Math · Mean Value Theorem and Extreme Value Theorem
AP Calculus AB: Mean Value Theorem and Extreme Value Theorem (Drill 1) is a Math practice drill covering Mean Value Theorem and Extreme Value Theorem. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice applying the Mean Value Theorem and Extreme Value Theorem to determine the existence of guaranteed values and absolute extrema on closed intervals. These AP Calculus AB topics appear regularly on both sections of the AP exam.
Question 1. Let \( f(x) = x^2 - 4x \) on the interval \( [1, 5] \). What value of \( c \) in \( (1, 5) \) is guaranteed by the Mean Value Theorem?
Explanation: Choice B is correct. The MVT guarantees a value \( c \in (1, 5) \) where \( f'(c) = \dfrac{f(5) - f(1)}{5 - 1} \). Computing: \( f(5) = 25 - 20 = 5 \) and \( f(1) = 1 - 4 = -3 \), so the average rate of change is \( \dfrac{5 - (-3)}{4} = \dfrac{8}{4} = 2 \). Since \( f'(x) = 2x - 4 \), setting \( 2c - 4 = 2 \) gives \( c = 3 \). Choice A is incorrect because \( f'(2) = 0 \neq 2 \); the student found where \( f'(x) = 0 \), a critical point, rather than where \( f'(c) \) equals the average rate of change. Choice C is incorrect because \( f'(4) = 4 \neq 2 \); this likely results from a setup error in which the student solved \( 2c - 4 = 4 \) rather than \( 2c - 4 = 2 \). Choice D is incorrect because \( f(x) = x^2 - 4x \) is a polynomial, which is continuous and differentiable everywhere; the MVT conditions are fully satisfied on \( [1, 5] \).
Question 2. The function \( h(x) = x^3 - 3x \) is continuous on \( [-2, 2] \). What are the absolute maximum and absolute minimum values of \( h \) on this interval?
Explanation: Choice A is correct. By the EVT, since \( h \) is continuous on a closed interval, it attains its absolute maximum and minimum. Finding critical points: \( h'(x) = 3x^2 - 3 = 0 \Rightarrow x = \pm 1 \). Evaluating at critical points and endpoints: \( h(-2) = -8 + 6 = -2 \); \( h(-1) = -1 + 3 = 2 \); \( h(1) = 1 - 3 = -2 \); \( h(2) = 8 - 6 = 2 \). The absolute maximum is 2 (attained at \( x = -1 \) and \( x = 2 \)) and the absolute minimum is \( -2 \) (attained at \( x = 1 \) and \( x = -2 \)). Choice B is incorrect because 0 is not a candidate value; this error occurs when a student evaluates \( h(0) = 0 \) (the y-intercept) instead of the critical points \( x = \pm 1 \) and endpoints \( x = \pm 2 \). Choice C is incorrect because \( h(-2) = (-2)^3 - 3(-2) = -8 + 6 = -2 \), not \( -4 \); the student made a sign error when cubing \( -2 \). Choice D is incorrect because \( h(2) = 8 - 6 = 2 \), not 4; the student likely computed \( 2^2 = 4 \) instead of \( 2^3 - 3(2) = 2 \).
Question 3. Which of the following functions satisfies the conditions of the Mean Value Theorem on \( [-1, 1] \)?
Explanation: Choice D is correct. The MVT requires the function to be continuous on \( [-1, 1] \) and differentiable on \( (-1, 1) \). The polynomial \( f(x) = x^3 + 2x \) is continuous and differentiable everywhere, so both conditions are satisfied. Choice A is incorrect because \( f(x) = |x| \) is not differentiable at \( x = 0 \); it has a corner there, and the derivative does not exist at that point, violating the differentiability condition. Choice B is incorrect because \( f(x) = \dfrac{1}{x} \) is undefined at \( x = 0 \in [-1, 1] \), so it is not continuous on \( [-1, 1] \); the continuity condition fails. Choice C is incorrect because \( \dfrac{d}{dx} x^{2/3} = \dfrac{2}{3} x^{-1/3} \) is undefined at \( x = 0 \), the graph has a cusp there, so the differentiability condition fails on \( (-1, 1) \).
Question 4. The table below shows selected values of a continuous function \( f \) on the closed interval \( [-1, 3] \).
| x | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| f(x) | 4 | −1 | 6 | 3 | −2 |
Explanation: Choice B is correct. The EVT guarantees that since \( f \) is continuous on the closed interval \( [-1, 3] \), it attains an absolute maximum. The table shows \( f(1) = 6 \), so the maximum is at least 6. However, since the table shows only selected values, \( f \) could attain values greater than 6 between the tabulated points; we cannot conclude the exact maximum from partial information alone. Choice A is incorrect because it asserts the maximum is exactly 6; the table does not show all values of \( f \), and a larger value could exist at a point not listed. Choice C is incorrect for the same reason: even though \( f(1) = 6 \) is the largest listed value, the absolute maximum could be attained at an unlisted point, so we cannot pin the location to \( x = 1 \). Choice D is incorrect because the table shows \( f(3) = -2 \) as the smallest listed value, but \( f \) could attain values less than \( -2 \) at unlisted points; the minimum is at most \( -2 \), not exactly \( -2 \).
Question 5. A car travels along a straight road. Its position (in miles) at time \( t \) (in hours) is given by a differentiable function \( s(t) \). If \( s(0) = 10 \) and \( s(4) = 90 \), which of the following must be true?
Explanation: Choice B is correct. Since \( s \) is differentiable on \( (0, 4) \) and continuous on \( [0, 4] \), the MVT guarantees the existence of some \( c \in (0, 4) \) where \( s'(c) = \dfrac{s(4) - s(0)}{4 - 0} = \dfrac{90 - 10}{4} = \dfrac{80}{4} = 20 \) mph. Since \( s'(t) \) represents velocity, the car’s velocity equals 20 mph at some moment in the open interval \( (0, 4) \). Choice A is incorrect because the average velocity is \( \dfrac{s(4) - s(0)}{4 - 0} = \dfrac{80}{4} = 20 \) mph, not 80 mph. The student confused the displacement (80 miles) with the average velocity, forgetting to divide by the elapsed time. Choice C is incorrect because the MVT guarantees the existence of some \( c \in (0, 4) \) where \( s'(c) = 20 \), but does not specify that \( c = 2 \); the guaranteed point could be anywhere in the open interval. Choice D is incorrect because the MVT makes no claim about the velocity at every point in the interval, the car could slow down, speed up, or even stop at various moments during \( (0, 4) \).