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

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Answer 1 · 2018-05-22T12:57:05

Critical points of $h^'(x)$ are $x=-9 and x =1$

$h^' (x) = x^2+8 x-9 =0$ or

$(x+9)(x-1)=0 :. x+9=0 or x =-9$ and

$x-1=0 :. x=1 :.$

Critical points of $h^'(x)$ are $x=-9 and x =1$ [Ans]

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