Home / AP Calculus AB 5.7 Using the Second Derivative Test to Determine Extrema – MCQs

AP Calculus AB 5.7 Using the Second Derivative Test to Determine Extrema - MCQs - Exam Style Questions

Calc-Ok Question

The function \(f\) is an antiderivative of the function \(g\) defined by \(g(x)=3-\sqrt{x^{2}+x+4\cos x}\). Which of the following is the \(x\)-coordinate of the location of a local maximum for the graph of \(y=f(x)\)?

(A) \(-3.961\)
(B) \(-2.161\)
(C) \(1.494\)
(D) \(3.140\)

▶️ Answer/Explanation
Key steps

\(f'(x)=g(x)\).
A local maximum of \(f\) occurs where \(g(x)=0\) and \(g'(x)<0\).

Set \(g(x)=0\): \(3-\sqrt{x^{2}+x+4\cos x}=0 \Rightarrow \sqrt{x^{2}+x+4\cos x}=3\).

Square: \(x^{2}+x+4\cos x-9=0\).

From the choices, zeros occur at \(x\approx -3.961\) and \(x\approx 3.140\).

Compute \(g'(x)=-\dfrac{2x+1-4\sin x}{2\sqrt{x^{2}+x+4\cos x}}\).

At a zero, \(\sqrt{\cdot}=3\Rightarrow\) denominator \(=6\).

At \(x\approx 3.140\): \(\sin x\approx 0\Rightarrow 2x+1-4\sin x>0\Rightarrow g'(x)<0\).
Then \(f”(x)=g'(x)<0\Rightarrow\) local maximum.

At \(x\approx -3.961\): \(2x+1-4\sin x<0\Rightarrow g'(x)>0\).
Then \(f”(x)=g'(x)>0\Rightarrow\) local minimum.

Therefore, the local maximum occurs at \(x=3.140\).
Answer: (D)

No-Calc Question

The function \(f\) given by \(f(x)=2x^{3}-3x^{2}-12x\) has a relative minimum at \(x=\)

(A) \(-1\)
(B) \(0\)
(C) \(2\)
(D) \(3-\dfrac{\sqrt{10}}{4}\)
(E) \(3+\dfrac{\sqrt{10}}{4}\)

▶️ Answer/Explanation

\(f'(x)=6x^{2}-6x-12=6(x-2)(x+1)\Rightarrow x=-1,2\).
\(f”(x)=12x-6\). Then \(f”(2)=18>0\Rightarrow\) relative minimum at \(x=2\).
Answer: (C)

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