AP Calculus AB Mean Value Theorem and Extreme Value Theorem Test Solutions and Scoring (PDF)

AP Calculus AB math test solutions: Mean Value Theorem and Extreme Value Theorem – Maths PDF

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This mvt evt test answers for AP Calculus AB covers mean value theorem and extreme value theorem (Common Core AP CALC AB 5.1, AP CALC AB 5.2, AP CALC AB 5.3), with a free PDF to print.

Use this answer key to score the mean value theorem test out of 20 points. Each solution shows the work and a scoring guide that says how the points are awarded. Give partial credit for a correct setup, a correct derivative or integral, and a correct final statement with units where they apply.

AP Calculus AB · Common Core State Standards: AP CALC AB 5.1, AP CALC AB 5.2, AP CALC AB 5.3

Mvt evt test answers: worked solutions

Question 1 – Verify the MVT [3 pts]

\(f\) is a polynomial, so it is continuous on \([0,4]\) and differentiable on \((0,4)\).

Average rate: \(\frac{f(4)-f(0)}{4}=\frac{8-0}{4}=2\). Solve \(2c-2=2\): \(c=2\).

Final answer: \(c=2\).

Graph of f(x) = x² − 2x.

Scoring: 1 point for the hypotheses; 1 point for the average rate; 1 point for c.

Question 2 – Hypotheses fail [2 pts]

\(f\) is undefined at \(x=2\), which is in the interval, so \(f\) is not continuous on \([1,3]\).

Scoring: 1 point for naming x = 2; 1 point for the conclusion about continuity.

Question 3 – MVT conclusion [2 pts]

\(g\) is differentiable, hence continuous. By the MVT there is a \(c\) with \(g^{\prime}(c)=\frac{-3-5}{5-1}=-2\).

Scoring: 1 point for the hypotheses; 1 point for the average rate -2.

Question 4 – Critical numbers [3 pts]

\(f^{\prime}=4x^{3}-16x=4x(x-2)(x+2)\): \(x=-2,\ 0,\ 2\).

\(h^{\prime}=\frac{2}{3}(x-1)^{-1/3}\) is never zero but undefined at \(x=1\).

Scoring: 1 point for f′; 1 point for the three numbers; 1 point for h.

Question 5 – Absolute extrema [3 pts]

\(f^{\prime}=3x^{2}-12=0\) gives \(x=\pm 2\). Candidates: \(f(-3)=12\), \(f(-2)=19\), \(f(2)=-13\), \(f(4)=19\).

Final answer: absolute maximum 19 (at x = -2 and x = 4); absolute minimum -13 (at x = 2).

Graph of f(x) = x³ − 12x + 3.

Scoring: 1 point for critical numbers; 1 point for the four values; 1 point for the conclusions.

Question 6 – Speed and the MVT [2 pts]

Position is continuous and differentiable. The average speed is \(\frac{150}{2.5}=60\) mph, so by the MVT the instantaneous speed equals \(60\) mph at some time.

Scoring: 1 point for the average speed; 1 point for citing the MVT.

Question 7 – Extrema with an exponential [3 pts]

\(f^{\prime}=e^{-x}(1-x)\), zero at \(x=1\). Candidates: \(f(0)=0\), \(f(1)=e^{-1}\approx 0.368\), \(f(5)=5e^{-5}\approx 0.034\).

Final answer: absolute maximum 1/e at x = 1; absolute minimum 0 at x = 0.

Scoring: 1 point for f′ and the critical number; 1 point for the values; 1 point for the conclusions.

Question 8 – Justifying a maximum [2 pts]

By the Extreme Value Theorem, \(h\) has an absolute maximum and minimum on \([1,6]\). They occur at the endpoints or the critical number.

The maximum is \(9\) at \(x=3\); the minimum is \(2\) at \(x=6\).

Scoring: 1 point for naming the EVT; 1 point for the two values with locations.

Total score

Total

20 points

Key facts behind this mvt evt test answers

Key facts

MVT: f′(c) = (f(b) – f(a))/(b – a) for some c in (a, b), if f is continuous and differentiable.
EVT: a continuous function on a closed interval has an absolute maximum and an absolute minimum.
Critical numbers are where f′ = 0 or f′ is undefined (inside the domain).
Candidates test: compare f at endpoints and critical numbers.

How to use this mvt evt test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

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