AP Calculus AB 5.7 Using the Second Derivative Test to Determine Extrema - MCQs - Exam Style Questions
Calc-Ok Question
(A) \(-3.961\)
(B) \(-2.161\)
(C) \(1.494\)
(D) \(3.140\)
▶️ Answer/Explanation
\(f'(x)=g(x)\).
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)