A) ? 4, 0
B) 0, ? 4
C) 4, 0
D) None of these
Correct Answer: B
Solution :
\[f'(x)=2(x-1)(x+2)+{{(x+2)}^{2}}=3{{x}^{2}}+6x\] \[f'(x)=0\] Þ\[x=0,\,-2\] \[f(-2)=(-2-1){{(-2+2)}^{2}}=0\] (Maximum value) and \[f(0)=(0-1){{(0+2)}^{2}}=-4\] (Minimum value).You need to login to perform this action.
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