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

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

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

critical points

\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

- sin (x) - cos (x)

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

y = x^{\frac{4}{3}} - x^{\frac{1}{3}}

2 x^{2} + ln (x)

f (x) = \frac{3 x ^{2}}{x ^{2} - 4}