Question

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

Answer 1

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

Explanation:

`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]