Question

Use the second derivative test to find local extrema of the function
$f\left(x\right)=-x^3+12x^2-5$ on $R$

Answer 1

$f\left(x\right)=-x^2+12x^2-5$

$f^{\prime }\left(x\right)=-3x^2+24x=0$

$x^2-8x=0$

$x(x-8)=0\Rightarrow x=0$ or $x=8$

$f^{\prime }\left(x\right)=-6x+24$

$f^{11}\left(x\right)=24>0f^{11}\left(8\right)=-48+24=-24<0$

$\Rightarrow x=0$ is minimum $x=8$ is maximum