A) 2
B) ?1
C) ? 2
D) 4
Correct Answer: D
Solution :
\[f'(x)=6{{x}^{2}}-6x-12\] \[f'(x)=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=-1,\,2\] Here \[f(4)=128-48-48+5=37\] \[f(-1)=-2-3+12+5=12\] \[f(2)=16-12-24+5=-15\] \[f(-2)=-16-12+24+5=1\] Therefore the maximum value of function is 37 at \[x=4\].You need to login to perform this action.
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