What is the critical (5x^2)/(x^2-9) ?

The critical (5x^2)/(x^2-9) is x=0

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critical (5x^2)/(x^2-9)

critical points

\frac{5 x ^{2}}{x ^{2} - 9}

x = 0

Solution steps

\frac{5 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{5 x ^{2}}{x ^{2} - 9} : x < - 3 or - 3 < x < 3 or x > 3

\frac{5 x ^{2}}{x ^{2} - 9}

is not defined at

x = - 3, x = 3

, therefore:

x = 0

g (x) = x^{3} - 26 x^{2}

f (x, y) = 1 - x^{2} - y^{2} + x y

\frac{6 e ^{x}}{6 e ^{x} + 4}

f (x) = 5 + x + \frac{5}{x}