Home / AP Calculus BC 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing – MCQs

AP Calculus BC 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing - MCQs - Exam Style Questions

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Question


The graph of \(f’\), the derivative of a function \(f\), is shown . Which of the following could be the graph of \(f\)?
▶️ Answer/Explanation
Detailed solution
\(f'(x)=-1\) for \(x<0\) ⟹ \(f\) is decreasing with slope \(-1\) on the left of \(0\).
\(f'(x)=1\) for \(x>0\) ⟹ \(f\) is increasing with slope \(+1\) on the right of \(0\).
The derivative jumps from \(-1\) to \(+1\) at \(x=0\), so \(f\) has a sharp corner (not smooth) at \(x=0\).

✅ Answer: (B)

Calc-Ok Question


The graph of a twice-differentiable function \(f\) is shown for \(0<x<4\). The graph has a horizontal tangent at \(x=2\) and points of inflection at \(x=1\) and \(x=3\). On what intervals do \(f'(x)\) and \(f”(x)\) have the same sign?
(A) \(0<x<1\) only
(B) \(2<x<4\)
(C) \(0<x<1\) and \(2<x<3\)
(D) \(1<x<2\) and \(3<x<4\)
▶️ Answer/Explanation
Detailed solution

Inflection points at \(x=1\) and \(x=3\) mean \(f”\) changes sign there. A horizontal tangent at \(x=2\) means \(f'(2)=0\) and the graph has a local minimum, so \(f’\) is negative on \((0,2)\) and positive on \((2,4)\).

Concavity: on \((0,1)\) the curve is concave down \((f”<0)\); on \((1,3)\) concave up \((f”>0)\); on \((3,4)\) concave down again.

Same sign occurs where:
• \((0,1)\): \(f'(x)<0\) and \(f”(x)<0\) ✔
• \((2,3)\): \(f'(x)>0\) and \(f”(x)>0\) ✔

Thus the intervals are \((0,1)\) and \((2,3)\).

✅ Answer: (C)

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