AP Calculus AB 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema - MCQs - Exam Style Questions
No-Calc Question
The first derivative of the function \(f\) is given by \(f'(x)=4x-8\). At what value of \(x\) does \(f\) attain an absolute maximum on the closed interval \([0,3]\) ?
(A) \(0\)
(B) \(\dfrac{3}{2}\)
(C) \(2\)
(D) \(3\)
▶️ Answer/Explanation
Critical point from \(f'(x)=0\): \(4x-8=0\Rightarrow x=2\).
Since \(f'(x)\) changes from negative to positive at \(x=2\), this is a minimum.
On \([0,3]\) the maximum occurs at an endpoint with larger \(f\)-value.
Because \(f'(x)>0\) for \(x>2\), \(f\) increases to \(x=3\).
✅ Answer: (D) \(x=3\)
No-Calc Question
What is the value of \(x\) at which the minimum value of \(y=3x^{4/3}-2x\) occurs on the closed interval \([0,1]\) ?
(A) \(0\)
(B) \(\tfrac{1}{8}\)
(C) \(\tfrac{1}{2}\)
(D) \(1\)
(B) \(\tfrac{1}{8}\)
(C) \(\tfrac{1}{2}\)
(D) \(1\)
▶️ Answer/Explanation
\(y’ = 4x^{1/3}-2\).
Set \(y’=0\): \(4x^{1/3}=2 \Rightarrow x^{1/3}=\tfrac12 \Rightarrow x=\tfrac18\).
Check endpoints: \(y(0)=0,\; y(1)=1,\; y(\tfrac18) = \tfrac{3}{16}-\tfrac14 = -\tfrac{1}{16}\).
Minimum occurs at \(x=\tfrac18\).
✅ Answer: (B)
Set \(y’=0\): \(4x^{1/3}=2 \Rightarrow x^{1/3}=\tfrac12 \Rightarrow x=\tfrac18\).
Check endpoints: \(y(0)=0,\; y(1)=1,\; y(\tfrac18) = \tfrac{3}{16}-\tfrac14 = -\tfrac{1}{16}\).
Minimum occurs at \(x=\tfrac18\).
✅ Answer: (B)