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

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

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

critical points

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

x = 0

Solution steps

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

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{2 x ^{2}}{x ^{2} - 9} : x < - 3 or - 3 < x < 3 or x > 3

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

is not defined at

x = - 3, x = 3

, therefore:

x = 0

f (x, y) = x^{2} - 4 x y + 2 y^{2} + 4 x + 8 y + 4

f (x) = \frac{e ^{4 x}}{x + 3}

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

f (x, y) = x^{3} - x + y^{3} - y

f (x) = x^{5} - x^{3}

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