Question

Which of the following is True about the function $f(x)=x^4-4x^2$ ?

• A. It has two local minima and one local maxima
• B. It has two local minima and zero local maxima
• C. It has one local minima and one local maxima
• D. It has two local minima and two local maxima

Answer 1

Answer: (A)

We get , $f\left(x\right)=x^4-4x^2$
$\Rightarrow f^{\prime }\left(x\right)=4x^3-8x$
$=4x\left(x^2-2\right)$
$\Rightarrow f^{\prime \prime }\left(x\right)=12x^2-8$
For maxima or minima, $f^{\prime }\left(x\right)=0$
$\Rightarrow 4x\left(x^2-2\right)=0$
$\Rightarrow x=0,\pm \sqrt{2}$
Now, ${\left[f^{\prime \prime }\left(x\right)\right]}_{x=0}=-8<0$
${\left[f^{\prime \prime }\left(x\right)\right]}_{x=-\sqrt{2}}=24-8=16>0$
and ${\left[f^{\prime \prime }\left(x\right)\right]}_{x=-\sqrt{2}}=24-8=16>0$
Thus, $x=0$ is a point of maxima and $x=\sqrt{2},-\sqrt{2}$ are points of minima