AP Calculus BC 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema - Exam Style Questions - MCQs - New Syllabus
Question
The first derivative of the function \(f\) is \( f'(x)=4x-8 \). 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) \(3\)
(B) \( \dfrac{3}{2} \)
(C) \(2\)
(D) \(3\)
▶️ Answer/Explanation
Detailed solution
Critical point: \(f'(x)=0\Rightarrow x=2\) (a local minimum since sign goes \( -\to +\)). Absolute extrema on \([0,3]\) occur at \(x=0,2,3\). Compare via the derivative: \[ f(3)-f(0)=\int_{0}^{3}(4x-8)\,dx=\big[2x^{2}-8x\big]_{0}^{3}=18-24=-6<0. \] Hence \(f(0)>f(3)\) and \(f(0)>f(2)\). ✅ Answer: (A) \(x=0\)
Critical point: \(f'(x)=0\Rightarrow x=2\) (a local minimum since sign goes \( -\to +\)). Absolute extrema on \([0,3]\) occur at \(x=0,2,3\). Compare via the derivative: \[ f(3)-f(0)=\int_{0}^{3}(4x-8)\,dx=\big[2x^{2}-8x\big]_{0}^{3}=18-24=-6<0. \] Hence \(f(0)>f(3)\) and \(f(0)>f(2)\). ✅ Answer: (A) \(x=0\)