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

The critical f(x)= x/(x^2+16) is x=-4,x=4

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critical f(x)= x/(x^2+16)

critical points

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

x = - 4, x = 4

Solution steps

\frac{x}{x ^{2} + 16}

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{x}{x ^{2} + 16} : - \infty < x < \infty

All critical points are in domain

x = - 4, x = 4

f (x) = \frac{2 + x}{2 - x}

7 x^{2} - 9 x - 5

sin (\frac{2}{x})

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

f (x) = \frac{3 x - 7}{x + 1}

What is the critical f(x)= x/(x^2+16) ?
The critical f(x)= x/(x^2+16) is x=-4,x=4