What is the critical (x^2)/(x^2-81) ?

The critical (x^2)/(x^2-81) is x=0

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critical (x^2)/(x^2-81)

critical points

\frac{x ^{2}}{x ^{2} - 81}

x = 0

Solution steps

\frac{x ^{2}}{x ^{2} - 81}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 9, x = 9

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{x ^{2} - 81} : x < - 9 or - 9 < x < 9 or x > 9

\frac{x ^{2}}{x ^{2} - 81}

is not defined at

x = - 9, x = 9

, therefore:

x = 0

f (x) = x^{2} e^{x}

f (x) = - x^{3} + 6 x^{2}

f (x) = 3 x - 24

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

\frac{1}{2} (x + 4)^{2} - 2