Question

The maximum of $f\left(x\right)=\frac{x^4}{4}-\frac{3}{2}{x}^2$ occurs at $x$

• A. $3$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $0$

Given: $f\left(x\right)=\frac{x^4}{4}-\frac{3}{2}{x}^2$

For maxima and minima
$f^{\prime }\left(x\right)=0$

$\Rightarrow x^3-3x=0$

$\Rightarrow x(x^2-3)=0$
$\Rightarrow x=0,\pm \sqrt{3}$
So these are the points of maxima and minima.
For maximum
$f^{\prime \prime }\left(x\right)<0$
So we will find $f^{\prime \prime }\left(x\right)$
$f^{\prime \prime }\left(x\right)=3x^2-3$
We can see that for $x=\pm \sqrt{3}$
$f^{\prime \prime }\left(x\right)=3x^2-3=6>0$
For $x=0$, $f^{\prime \prime }\left(x\right)=3x^2-3=-3<0$
So maximum of $f\left(x\right)$ occurs at $x=0$ .