Home / AP Calculus BC 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema – MCQs

AP Calculus BC 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema - MCQs - Exam Style Questions

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Question

Let \(f\) be twice differentiable. Selected values of \(f’\) (the first derivative of \(f\)) are given in the table. On which of the following intervals is \(f\) guaranteed to attain a local maximum?

\(x\)012345
\(f'(x)\)1\(-2\)3050

(A) \((0,1)\)
(B) \((1,2)\)
(C) \((2,4)\)
(D) \((3,5)\)

▶️ Answer/Explanation
Detailed solution

Since \(f\) is twice differentiable, \(f’\) is continuous on the intervals considered.

A local maximum occurs where \(f’\) changes sign from \(+\) to \(-\).

• On \((0,1)\): \(f'(0)=1>0\) and \(f'(1)=-2<0\). By the Intermediate Value Theorem for continuous \(f’\), there exists \(c\in(0,1)\) with \(f'(c)=0\) and the sign changes \(+\to-\) ⇒ local maximum.

• On \((1,2)\): \(-2\to 3\) is a sign change \(-\to+\) ⇒ local minimum (not a maximum).

• On \((2,4)\): \(3\to 0\to 5\) stays nonnegative ⇒ no guaranteed local max.

• On \((3,5)\): \(0\to 5\to 0\) shows nonnegative values; a \(+\to-\) change is not guaranteed.

✅ Answer: (A) \((0,1)\)

Calc-Ok Question

If \(f'(x)=1+2x-x^{3}-\sqrt{x^{4}+1}\), then \(f\) has a local minimum at \(x=\)
(A) \(-1.836\)
(B) \(-1.117\)
(C) \(0\)
(D) \(1.181\)
▶️ Answer/Explanation
Detailed solution

Critical points solve \(f'(x)=0\). At \(x=0\): \(f'(0)=1+0-0-\sqrt{1}=0\).

Check sign: for small negative \(x\), \(f'(x)\approx 2x – x^{3} – \tfrac{1}{2}x^{4} < 0\); for small positive \(x\), \(f'(x)>0\). Thus \(f’\) changes from negative to positive at \(x=0\) ⇒ local minimum.

✅ Answer: (C)

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