This ap calculus analyzing test answer key gives the full solution and the scoring guide for each of the 8 test questions on analyzing functions with derivatives.
Worked solutions and scoring guides for the Unit 5 test. Each answer shows the reasoning an AP reader looks for: the derivative, the sign analysis or the named theorem, and a clear conclusion. Award points exactly as the scoring note describes, and total them out of 20.
AP Calculus Analyzing Test Answer Key: questions 1–5
Question 1: Critical numbers and monotonicity (2 points)
(a) \(f^{\prime}(x)=3x^{2}-6x-24=3(x-4)(x+2)\), so the critical numbers are \(x=-2\) and \(x=4\).
(b) \(f^{\prime} > 0\) on \((-\infty,-2)\) and \((4,\infty)\); \(f^{\prime} < 0\) on \((-2,4)\). So \(f\) is increasing on \((-\infty,-2)\) and \((4,\infty)\) and decreasing on \((-2,4)\). \(f^{\prime}\) changes from \(+\) to \(-\) at \(x=-2\) (local maximum \(f(-2)=33\)) and from \(-\) to \(+\) at \(x=4\) (local minimum \(f(4)=-75\)).
Scoring: (a) 1 point for both critical numbers. (b) 1 point for the correct intervals and the classification with the sign-change reason.
Question 2: Mean Value Theorem (2 points)
(a) \(f\) is a polynomial, so it is continuous on \([0,2]\) and differentiable on \((0,2)\).
(b) \(\frac{f(2)-f(0)}{2-0}=\frac{6-0}{2}=3\). Solving \(3c^{2}-1=3\) gives \(c^{2}=\frac{4}{3}\), so \(c=\frac{2\sqrt{3}}{3}\approx 1.155\) (the negative root is outside the interval).
Scoring: (a) 1 point for naming both continuity and differentiability. (b) 1 point for the correct value of \(c\) with the root in \((0,2)\) chosen.
Question 3: Reading the graph of f prime (2 points)
(a) \(f^{\prime}(x)=x^{2}-2x-3=(x-3)(x+1)\). \(f^{\prime}\) changes from positive to negative at \(x=-1\), so \(f\) has a local maximum at \(x=-1\); it changes from negative to positive at \(x=3\), so \(f\) has a local minimum at \(x=3\).
(b) \(f^{\prime}\) is decreasing for \(x < 1\) and increasing for \(x > 1\), so \(f^{\prime\prime}\) changes from negative to positive at \(x=1\). The inflection point is at \(x=1\).
Scoring: (a) 1 point for \(x=-1\) (maximum) and \(x=3\) (minimum) with the sign-change reason. (b) 1 point for \(x=1\) with the reason that \(f^{\prime}\) changes from decreasing to increasing.
Question 4: Candidates Test (3 points)
\(f^{\prime}(x)=4x^{3}-16x=4x(x-2)(x+2)\). The critical numbers in \([-1,3]\) are \(x=0\) and \(x=2\) (\(x=-2\) is outside).
\(f(-1)=-4\), \(f(0)=3\), \(f(2)=-13\), \(f(3)=12\).
The absolute maximum is \(12\) (at \(x=3\)) and the absolute minimum is \(-13\) (at \(x=2\)).
Scoring: 1 point for the critical numbers in the interval; 1 point for the four function values; 1 point for the correct maximum and minimum.
Question 5: Inscribed rectangle (3 points)
With upper right vertex \((x,12-x^{2})\), the width is \(2x\) and the height is \(12-x^{2}\), so \(A(x)=2x(12-x^{2})=24x-2x^{3}\) for \(0 < x < 2\sqrt{3}\).
\(A^{\prime}(x)=24-6x^{2}=0\) gives \(x=2\) (positive root). \(A^{\prime\prime}(x)=-12x\), so \(A^{\prime\prime}(2)=-24 < 0\): a maximum. (Also, \(A\to 0\) at both ends of the domain.)
Width \(=4\), height \(=12-4=8\), area \(=32\).
Scoring: 1 point for \(A(x)\) and its derivative; 1 point for \(x=2\); 1 point for the justification of the maximum and the dimensions \(4\times 8\) with area \(32\).
AP Calculus Analyzing Test Answer Key: questions 6–8
Question 6: Implicit curve (3 points)
(a) Differentiate: \(2x-y-x\frac{dy}{dx}+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=\frac{y-2x}{2y-x}\).
(b) At \((1,2)\): \(\frac{dy}{dx}=\frac{2-2}{4-1}=0\), so the tangent is horizontal.
(c) \(\frac{d^{2}y}{dx^{2}}=\frac{(\frac{dy}{dx}-2)(2y-x)-(y-2x)(2\frac{dy}{dx}-1)}{(2y-x)^{2}}\). At \((1,2)\) with \(\frac{dy}{dx}=0\): \(\frac{(-2)(3)-0}{9}=-\frac{2}{3}\). Since it is negative, the curve is concave down and has a local maximum at \((1,2)\).
Scoring: (a) 1 point for the correct derivative. (b) 1 point for substituting and getting 0. (c) 1 point for \(-\frac{2}{3}\) and the conclusion (local maximum).
Question 7: Factored derivative (2 points)
(a) The factor \((x-4)^{2}\) is nonnegative, so the sign of \(g^{\prime}\) matches the sign of \(x-1\) (for \(x\neq 4\)). \(g^{\prime} > 0\) for \(x > 1\) with \(x\neq 4\), and since \(g\) is continuous, \(g\) is increasing on \((1,\infty)\).
(b) No. \(g^{\prime}(4)=0\), but \(g^{\prime}\) is positive on both sides of \(x=4\), so there is no sign change and no local extremum.
Scoring: (a) 1 point for \((1,\infty)\). (b) 1 point for “no” with the reason that \(g^{\prime}\) does not change sign.
Question 8: Open-top storage bin (3 points)
\(x^{2}h=32\) gives \(h=\frac{32}{x^{2}}\). The surface area is \(S=x^{2}+4xh=x^{2}+\frac{128}{x}\) for \(x > 0\).
\(S^{\prime}(x)=2x-\frac{128}{x^{2}}=0\) gives \(x^{3}=64\), so \(x=4\). \(S^{\prime\prime}(x)=2+\frac{256}{x^{3}} > 0\), so \(x=4\) gives a minimum.
\(h=\frac{32}{16}=2\). \(S(4)=16+32=48\). The bin is 4 ft by 4 ft by 2 ft, with surface area 48 square feet.
Scoring: 1 point for \(S(x)=x^{2}+\frac{128}{x}\); 1 point for \(S^{\prime}\) and \(x=4\); 1 point for the minimum justification, \(h=2\) and \(S=48\).
Scoring summary
20 points: add the points of every question of this ap calculus analyzing test answer key.
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Use this ap calculus analyzing test answer key to score the test fairly and to see which skills need more practice.
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AP Calculus AB Analyzing Functions with Derivatives test answer key
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