Home / AP Calculus AB 5.6 Determining Concavity of Functions over Their Domains – MCQs

AP Calculus AB 5.6 Determining Concavity of Functions over Their Domains - MCQs - Exam Style Questions

AP Calculus AB-MCQs and Answers - All Topics

No-Calc Question

A store sold an item over a 16-month period. The total number of items sold up to time \(t\) is modeled by a twice-differentiable function \(S(t)\), where \(t\) is the number of months since the store started selling the item. For the first 8 months, \(S’\) is increasing. After 8 months, \(S’\) is decreasing. Which of the following could be the graph of \(S(t)\)?

▶️ Answer/Explanation

\(S’\) increasing on \([0,8]\) ⇒ \(S”>0\) (concave up).
\(S’\) decreasing after \(8\) ⇒ \(S”<0\) (concave down).
\(S\) is always increasing (counts items).
The only sketch showing “concave up then concave down” with monotonic increase is option **(C)**.

✅ Answer: (C)

Calc-Ok Question

The function \( f \) has first derivative \( f’ \) defined by \( f'(x)=x^{2}\tan(1-2x) \) for \(-0.25<x<0.75\). At which of the following values of \( x \) does the graph of \( f \) have a point of inflection?

(A) \( x=0.324 \) only
(B) \( x=0.5 \) only
(C) \( x=0 \) and \( x=0.324 \)
(D) \( x=0 \) and \( x=0.5 \)

▶️ Answer/Explanation
Calculation steps

Compute \(f”(x)\).
Use product rule on \(x^{2}\) and \(\tan(1-2x)\).
Use chain rule: \(\dfrac{d}{dx}\tan(1-2x)=\sec^{2}(1-2x)\cdot(-2)\).
So \(f”(x)=2x\,\tan(1-2x)-2x^{2}\sec^{2}(1-2x)\).
Factor: \(f”(x)=2x\big[\tan(1-2x)-x\sec^{2}(1-2x)\big]\).

Set \(f”(x)=0\).
Either \(2x=0 \Rightarrow x=0\).
Or \(\tan(1-2x)=x\sec^{2}(1-2x)\).

Solve \(\tan(1-2x)=x\sec^{2}(1-2x)\) numerically on \((-0.25,0.75)\).
Root occurs at \(x\approx 0.324\).

Check sign change of \(f”\).
Near \(x=0\): \(f”(-0.1)<0\) and \(f”(0.1)>0\) ⇒ sign changes.
Near \(x=0.324\): \(f”(0.323)>0\) and \(f”(0.326)<0\) ⇒ sign changes.

✅ Answer: (C) \(x=0\) and \(x=0.324\)

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