What is the critical f(x)=(6x^2)/(x^2-25) ?

The critical f(x)=(6x^2)/(x^2-25) is x=0

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critical f(x)=(6x^2)/(x^2-25)

critical points

f (x) = \frac{6 x ^{2}}{x ^{2} - 25}

x = 0

Solution steps

\frac{6 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{6 x ^{2}}{x ^{2} - 25} : x < - 5 or - 5 < x < 5 or x > 5

\frac{6 x ^{2}}{x ^{2} - 25}

is not defined at

x = - 5, x = 5

, therefore:

x = 0

\frac{5 x}{x ^{2} - 36}

- x^{5} - 5 x^{4}

\frac{2 x ^{2}}{x ^{2} - 16}

f (x) = sin (y)

f (x) = 2 x^{3} - 3 x^{2} - 36 x + 14

What is the critical f(x)=(6x^2)/(x^2-25) ?
The critical f(x)=(6x^2)/(x^2-25) is x=0