What is the critical f(x)=(x^2)/(x^2-36) ?

The critical f(x)=(x^2)/(x^2-36) is x=0

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critical f(x)=(x^2)/(x^2-36)

critical points

f (x) = \frac{x ^{2}}{x ^{2} - 36}

x = 0

Solution steps

\frac{x ^{2}}{x ^{2} - 36}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 6, x = 6

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{x ^{2} - 36} : x < - 6 or - 6 < x < 6 or x > 6

\frac{x ^{2}}{x ^{2} - 36}

is not defined at

x = - 6, x = 6

, therefore:

x = 0

f (x) = \frac{24}{x ^{2} + 12}

x^{2} + x y + 3 x + 2 y + 5

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

f (x) = 2 (x - 4)^{2} (y^{2} + 3) + 3 y^{2} + 12 y

x^{2} - 2

What is the critical f(x)=(x^2)/(x^2-36) ?
The critical f(x)=(x^2)/(x^2-36) is x=0