Question

Find the critical points of the following function and test them for their maxima and minima.
$y=\frac{1}{5}{x}^5-4x^2$

Answer 1

$y=\frac{1}{5}{x}^5-4x^2$

$\frac{dy}{dx}=x^4-8x=0$

$x(x^3-8)=0$

$x=0;x=2$

$\frac{d^{2}y}{d{x}^2}=4x^3-8$

for $x=0,4x^3-8<0$ Hence $x=0$ is a maximum for this function

For $x=2,4x^3-8>0$ Hence $x=2$ is a minimum for this function