Question

Let $f:[-1,1]\to R,$ where $f\left(x\right)=2x^3-x^4-10.$ The minimum value of $f\left(x\right)$ is

• A. -13

Answer 1

The correct option is A -13
$f\left(x\right)=2x^3-x^4-10,$
$f^{\prime }\left(x\right)=6x^2-4x^3,$
$f^{\prime \prime }\left(x\right)=12x-12x^2,$
$f^{\prime \prime \prime }\left(x\right)=12-24x$
For $CriticalPoints;f^{\prime }\left(x\right)=0$
$\Rightarrow 6x^2-4x^3=0$
$\Rightarrow x=0,x=\frac{3}{2}$ are stationary points
$x=\frac{3}{2}\text{/}ϵ[-1,1]$ So we have only one point $i.e.,x=0$
$\because f^{\prime \prime }\left(0\right)=0$ and $f^{\prime \prime }\left(0\right)\ne 0$
So $x=0$ is a point of inflection
So Minimum value occur only at corner points
$i.e.,f(-1)=-13\&f\left(1\right)=-9$
So Min. $f\left(x\right)=-13$.