What is the critical f(x)=((x+4))/(x^2) ?

The critical f(x)=((x+4))/(x^2) is x=-8

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critical f(x)=((x+4))/(x^2)

critical points

f (x) = \frac{( x + 4 )}{x ^{2}}

x = - 8

Solution steps

\frac{x + 4}{x ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 8, x = 0

Identify critical points not in

f (x)

domain

Domain of

\frac{x + 4}{x ^{2}} : x < 0 or x > 0

\frac{x + 4}{x ^{2}}

is not defined at

x = 0

, therefore:

x = - 8

f (x) = 4 x^{3} - 12 x^{2} - 9 x + 27

f (x) = \frac{x}{7 x - 4}

6 x - 2

\frac{9}{5}

f (x) = ln (10 - x)

What is the critical f(x)=((x+4))/(x^2) ?
The critical f(x)=((x+4))/(x^2) is x=-8