AP Calculus AB 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing - MCQs - Exam Style Questions
Calc-Ok Question
The function \(f\) is continuous on the closed interval \([0,5]\). The graph of \(f’\), the derivative of \(f\), is shown above. On which of the following intervals is \(f\) increasing?
(A) \([0,1]\) and \([2,4]\)
(B) \([0,1]\) and \([3,5]\)
(C) \([0,1]\) and \([4,5]\) only
(D) \([0,2]\) and \([4,5]\)
(B) \([0,1]\) and \([3,5]\)
(C) \([0,1]\) and \([4,5]\) only
(D) \([0,2]\) and \([4,5]\)
▶️ Answer/Explanation
\(f\) increases where \(f'(x)>0\).
From the graph, \(f'(x)>0\) on roughly \([0,2]\) and \([4,5]\).
✅ Answer: (D)
From the graph, \(f'(x)>0\) on roughly \([0,2]\) and \([4,5]\).
✅ Answer: (D)
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)