What is the critical f(x)=(2x-3)/(x^2-9) ?

The critical f(x)=(2x-3)/(x^2-9) is None

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critical f(x)=(2x-3)/(x^2-9)

critical points

f (x) = \frac{2 x - 3}{x ^{2} - 9}

None

Solution steps

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

\frac{2 x - 3}{x ^{2} - 9}

is not defined at

x = - 3, x = 3

, therefore:

None

f (x) = (x^{2} - 1) e^{- a x}

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

5 x^{\frac{2}{3}} - 2 x^{\frac{5}{3}}

\frac{p - 1}{p ^{2} + 4}

3 x^{5} - 5 x^{3}

What is the critical f(x)=(2x-3)/(x^2-9) ?
The critical f(x)=(2x-3)/(x^2-9) is None