AP Calculus AB 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points - MCQs - Exam Style Questions
No-Calc Question
Let \(f(x)=k\sqrt{x}+\dfrac{1}{x}\) for \(x>0\), where \(k\) is a constant. If \(f\) has a relative extremum at \(x=9\), what is \(k\)?
(A) \(\dfrac{2}{3}\)
(B) \(\dfrac{2}{27}\)
(C) \(\dfrac{2}{81}\)
(D) \(\dfrac{2}{243}\)
(B) \(\dfrac{2}{27}\)
(C) \(\dfrac{2}{81}\)
(D) \(\dfrac{2}{243}\)
▶️ Answer/Explanation
Set \(f'(9)=0\). Compute \(f'(x)=\dfrac{k}{2\sqrt{x}}-\dfrac{1}{x^{2}}\).
Then \(0=f'(9)=\dfrac{k}{2\cdot 3}-\dfrac{1}{9^{2}}=\dfrac{k}{6}-\dfrac{1}{81}\).
Solve: \(\dfrac{k}{6}=\dfrac{1}{81}\Rightarrow k=\dfrac{2}{27}\).
✅ Answer: (B)
Then \(0=f'(9)=\dfrac{k}{2\cdot 3}-\dfrac{1}{9^{2}}=\dfrac{k}{6}-\dfrac{1}{81}\).
Solve: \(\dfrac{k}{6}=\dfrac{1}{81}\Rightarrow k=\dfrac{2}{27}\).
✅ Answer: (B)