A) 0
B) 12
C) 16
D) 32
Correct Answer: B
Solution :
\[y=f(x)=-{{x}^{3}}+3{{x}^{2}}+9x-27\] The slope of this curve \[{f}'(x)=-3{{x}^{2}}+6x+9\] Let \[g(x)={f}'(x)=-3{{x}^{2}}+6x+9\] Differentiate with respect to x, \[{g}'(x)=-6x+6\] Put \[{g}'(x)=0\] Þ \[x=1\] Now, \[{g}''(x)=-6<0\] and hence at \[x=1,\] \[g(x)\] (slope) will have maximum value. \ \[{{[g(1)]}_{\text{max}\text{.}}}=-3\times 1+6+9=12\].You need to login to perform this action.
You will be redirected in
3 sec