MHT CET · Maths · Application of Derivatives
Local maximum and local minimum values respectively of the function \(f(x)=(x-1)(x+2)^2\) are
- A \(-4,0\)
- B \(0,-4\)
- C \(-4,4\)
- D \(4,-4\)
Answer & Solution
Correct Answer
(B) \(0,-4\)
Step-by-step Solution
Detailed explanation
\(f(x)=(x-1)(x+2)^2 \)
\( f^{\prime}(x)=(x+2)^2+(x-1) 2(x+2)=(x+2)\)\((x+2+2 x-2)=3 x(x+2) \)
\( \stackrel{+}{+}-0_{-2}^{+} \)
\( f_{\max }=f(-2)=0 \)
\( f_{\min }=f(0)=-4\)
\( f^{\prime}(x)=(x+2)^2+(x-1) 2(x+2)=(x+2)\)\((x+2+2 x-2)=3 x(x+2) \)
\( \stackrel{+}{+}-0_{-2}^{+} \)
\( f_{\max }=f(-2)=0 \)
\( f_{\min }=f(0)=-4\)
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