How do you find the critical points of h'(x)=x^2+8x-9?

May 22, 2018

Critical points of ${h}^{'} \left(x\right)$ are $x = - 9 \mathmr{and} x = 1$

Explanation:

${h}^{'} \left(x\right) = {x}^{2} + 8 x - 9 = 0$ or

$\left(x + 9\right) \left(x - 1\right) = 0 \therefore x + 9 = 0 \mathmr{and} x = - 9$ and

$x - 1 = 0 \therefore x = 1 \therefore$

Critical points of ${h}^{'} \left(x\right)$ are $x = - 9 \mathmr{and} x = 1$ [Ans]