What is the critical (-x^2-9)/((x^2-9)^2) ?

The critical (-x^2-9)/((x^2-9)^2) is x=0

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critical (-x^2-9)/((x^2-9)^2)

critical points

\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}}

x = 0

Solution steps

\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 3, x = 3

Identify critical points not in

f (x)

domain

Domain of

\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}} : x < - 3 or - 3 < x < 3 or x > 3

\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}}

is not defined at

x = - 3, x = 3

, therefore:

x = 0

x^{5} + 5 x^{4} - 40 x^{2}

f (x) = 4 x^{4} - 4 x^{3}

- 2^{x}

f (x) = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} - 10 x

90 x^{2} e^{- 1.2 x}

What is the critical (-x^2-9)/((x^2-9)^2) ?
The critical (-x^2-9)/((x^2-9)^2) is x=0