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

1 Answer

May 22, 2018

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]

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