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Question: Answered & Verified by Expert
Local maximum and local minimum values respectively of the function $f(x)=(x-1)(x+2)^2$ are
MathematicsApplication of DerivativesJEE Main
Options:
  • A $-4,0$
  • B $0,-4$
  • C $-4,4$
  • D $4,-4$
Solution:
1123 Upvotes Verified Answer
The correct answer is: $0,-4$
$\begin{aligned} & f(x)=(x-1)(x+2)^2 \\ & f^{\prime}(x)=(x+2)^2+(x-1) 2(x+2)=(x+2)(x+2+2 x-2)=3 x(x+2) \\ & \stackrel{+}{+}-0_{-2}^{+} \\ & f_{\max }=f(-2)=0 \\ & f_{\min }=f(0)=-4\end{aligned}$

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