What is the critical f(x)=(x^2-7)/(16-x^2) ?

The critical f(x)=(x^2-7)/(16-x^2) is x=0

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critical f(x)=(x^2-7)/(16-x^2)

critical points

f (x) = \frac{x ^{2} - 7}{16 - x ^{2}}

x = 0

Solution steps

\frac{x ^{2} - 7}{16 - x ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 4, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} - 7}{16 - x ^{2}} : x < - 4 or - 4 < x < 4 or x > 4

\frac{x ^{2} - 7}{16 - x ^{2}}

is not defined at

x = - 4, x = 4

, therefore:

x = 0

f (x) = 2 x^{2} - 5 x + 1

- 9 - x^{2}

- 2 (x - 3)^{2} + 7

f (x) = \frac{x ^{5} + 10}{3}

f (x) = 2 x - 4

What is the critical f(x)=(x^2-7)/(16-x^2) ?
The critical f(x)=(x^2-7)/(16-x^2) is x=0