Question
Find the maximum and minimum values of $f(x)=2 x^{3}+3 x^{2}-12 x-4$
✓
Answer
Here, $f(x)=2 x^{3}+3 x^{2}-12 x-4$
$ \therefore f^{\prime}(x)=6 x^{2}+6 x-12 $
For stationary values, $f^{\prime}(x)=0$
$ \therefore 6 x^{2}+6 x-12=0$
$\therefore x^{2}+x-2=0$
$\therefore (x+2)(x-1)=0$
$\therefore x=-2 \text { or } x=1 $
Now, $f^{\prime \prime}(x)=12 x+6$
$ \text { At } \boldsymbol{x}=-\mathbf{2} $
$f^{\prime \prime}(-2) =12(-2)+6$
$ =-18<0 $
$\therefore $ We get the maximum value of the function at $x=-2$.
At $x=1$
$ f^{\prime \prime}(1) =12(1)+6$
$ =18>0 $
$\therefore$ We get the minimum value of the function at $x=1$.
Minimum value of $f(x)$
Putting $x=1$ in the function $f(x)$,
$ f(1) =2(1)^{3}+3(1)^{2}-12(1)-4$
$ =2+3-12-4$
$ =-11 $
Maximum value of $f(x)$
Putting $x=-2$ in the function $f(x)$,
$ f(-2) =2(-2)^{3}+3(-2)^{2}-12(-2)-4$
$ =-16+12+24-4$
$ =16 $
Thus, the maximum value of $f(x)$ is $16$ and the minimum value is $-11$.
$ \therefore f^{\prime}(x)=6 x^{2}+6 x-12 $
For stationary values, $f^{\prime}(x)=0$
$ \therefore 6 x^{2}+6 x-12=0$
$\therefore x^{2}+x-2=0$
$\therefore (x+2)(x-1)=0$
$\therefore x=-2 \text { or } x=1 $
Now, $f^{\prime \prime}(x)=12 x+6$
$ \text { At } \boldsymbol{x}=-\mathbf{2} $
$f^{\prime \prime}(-2) =12(-2)+6$
$ =-18<0 $
$\therefore $ We get the maximum value of the function at $x=-2$.
At $x=1$
$ f^{\prime \prime}(1) =12(1)+6$
$ =18>0 $
$\therefore$ We get the minimum value of the function at $x=1$.
Minimum value of $f(x)$
Putting $x=1$ in the function $f(x)$,
$ f(1) =2(1)^{3}+3(1)^{2}-12(1)-4$
$ =2+3-12-4$
$ =-11 $
Maximum value of $f(x)$
Putting $x=-2$ in the function $f(x)$,
$ f(-2) =2(-2)^{3}+3(-2)^{2}-12(-2)-4$
$ =-16+12+24-4$
$ =16 $
Thus, the maximum value of $f(x)$ is $16$ and the minimum value is $-11$.
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