Question

Find the greatest and the least values of the following functions.
$y=x^4-8x^2-9$ on the interval [-1, 1].

Answer 1

$y=x^4-8x^2-9$

$\frac{dy}{dx}=4x^3-16x$

$x=-2,0,2$

$\frac{d^{2}y}{d{x}^2}=12x^2-16$

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=-2,2}=32>0$,Hence $f(-2)=f\left(20\right)=-25$ is the minimum

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=0}=-16<0$,Hence $f\left(0\right)=-9$ is the maximum

$y_{x=1}=-16$ is the minimum in the interval $[-1,1]$