Question

Find the critical points of the following function and test them for their maxima and minima.
$y=x^5{x}^2+8$

Answer 1

$y=x^5-x^2+8$

$\frac{dy}{dx}=5x^4-2x=0$

$x=0,(\frac{2}{5}{)}^{\frac{1}{3}}$

$\frac{d^{2}y}{d{x}^2}=20x^3-2$

For $x=0,\frac{d^{2}y}{d{x}^2}<0$ Hence it is a maximum.

For $x=(\frac{2}{5}{)}^{\frac{1}{3}},\frac{d^{2}y}{d{x}^2}>0$,Hence it is a minimum.