What is the critical (2x^2)/(x^2-25) ?

The critical (2x^2)/(x^2-25) is x=0

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critical (2x^2)/(x^2-25)

critical points

\frac{2 x ^{2}}{x ^{2} - 25}

x = 0

Solution steps

\frac{2 x ^{2}}{x ^{2} - 25}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 5, x = 5

Identify critical points not in

f (x)

domain

Domain of

\frac{2 x ^{2}}{x ^{2} - 25} : x < - 5 or - 5 < x < 5 or x > 5

\frac{2 x ^{2}}{x ^{2} - 25}

is not defined at

x = - 5, x = 5

, therefore:

x = 0

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

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

f (x) = 2 sin (\frac{x}{2})

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

f (x) = x^{3} - 3 x^{2} + 4 x + 5