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

The critical f(x)= x/(sqrt(x^2-9)) is None

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

critical points

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

None

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 3

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = - 3, x = 3

, therefore:

None

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

f (x, y) = x^{4} + y^{4} + 6 x^{2} y^{2} + 8 x^{3}

x^{4} - 4 x^{3} + 7

f (x) = \frac{5 x}{x - 3}

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

What is the critical f(x)= x/(sqrt(x^2-9)) ?
The critical f(x)= x/(sqrt(x^2-9)) is None