AP Calculus AB 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing - MCQs - Exam Style Questions
No-Calc Question
The function \(f\) has derivative \(f'(x)=3x^{2}+x-4\). On which interval is the graph of \(f\) both decreasing and concave up?
(A) \((-\infty,-\tfrac{1}{6})\)
(B) \((-\tfrac{4}{3},-\tfrac{1}{6})\) only
(C) \((-\tfrac{1}{6},1)\) only
(D) \((-\tfrac{1}{6},\infty)\)
(B) \((-\tfrac{4}{3},-\tfrac{1}{6})\) only
(C) \((-\tfrac{1}{6},1)\) only
(D) \((-\tfrac{1}{6},\infty)\)
▶️ Answer/Explanation
Decreasing when \(f'(x)<0\). Factor: \(3x^{2}+x-4=(x+1)(3x+4)\).
Roots at \(x=-\tfrac{4}{3}\) and \(x=1\); so \(f'(x)<0\) on \((-\tfrac{4}{3},1)\).
Concave up when \(f”(x)=6x+1>0\Rightarrow x>-\tfrac{1}{6}\).
Intersection: \((-\tfrac{1}{6},1)\).
✅ Answer: (C)
Roots at \(x=-\tfrac{4}{3}\) and \(x=1\); so \(f'(x)<0\) on \((-\tfrac{4}{3},1)\).
Concave up when \(f”(x)=6x+1>0\Rightarrow x>-\tfrac{1}{6}\).
Intersection: \((-\tfrac{1}{6},1)\).
✅ Answer: (C)
Calc-Ok Question
The graph of \(f’\) (the derivative of \(f\)) is shown for \(-2<x<4\). Which statement about \(f\) must be true on this interval?
(A) one relative extremum; one inflection point
(B) one relative extremum; two inflection points
(C) two relative extrema; one inflection point
(D) two relative extrema; two inflection points
(B) one relative extremum; two inflection points
(C) two relative extrema; one inflection point
(D) two relative extrema; two inflection points
▶️ Answer/Explanation
Relative extrema of \(f\) occur where \(f’\) crosses \(0\) with sign change — the graph crosses once.
Inflection points of \(f\) occur where \(f’\) has a local extremum — the graph of \(f’\) shows two turning points.
✅ Answer: (B)
Inflection points of \(f\) occur where \(f’\) has a local extremum — the graph of \(f’\) shows two turning points.
✅ Answer: (B)