Question

Find the critical points of the following functions.
$y=\frac{x^3}{3}+2x^2-5x+4$

Answer 1

$y=\frac{x^3}{3}+2x^2-5x+4$

$⟹\frac{dy}{dx}=x^2+4x-5$

For critical points-

$\frac{dy}{dx}=0$

$⟹x^2+4x-5=0$

$⟹(x+5)(x-1)=0$

$⟹x=-5,1$

Hence, critical points are $x=-5$ and $x=1$