What is the critical (9x)/(16-x^2) ?

The critical (9x)/(16-x^2) is None

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critical (9x)/(16-x^2)

critical points

\frac{9 x}{16 - x ^{2}}

None

Solution steps

\frac{9 x}{16 - x ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{9 x}{16 - x ^{2}} : x < - 4 or - 4 < x < 4 or x > 4

\frac{9 x}{16 - x ^{2}}

is not defined at

x = - 4, x = 4

, therefore:

None

in v erse f (x) = (x^{7} - 9)^{9}

in f l ec t i o n (x - 4)^{3}

e x t re m e (x - 5)^{\frac{2}{3}}

p a r i t y f (x) = \frac{tan ( x )}{4 - x ^{2}}

d o main f (x) = 0. 5^{x}