What is the critical f(x)= 1/(x+3)-1/((x+3)^2) ?

The critical f(x)= 1/(x+3)-1/((x+3)^2) is x=-1

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critical f(x)= 1/(x+3)-1/((x+3)^2)

critical points

f (x) = \frac{1}{x + 3} - \frac{1}{( x + 3 ) ^{2}}

x = - 1

Solution steps

\frac{1}{x + 3} - \frac{1}{( x + 3 ) ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = - 3

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{x + 3} - \frac{1}{( x + 3 ) ^{2}} : x < - 3 or x > - 3

\frac{1}{x + 3} - \frac{1}{( x + 3 ) ^{2}}

is not defined at

x = - 3

, therefore:

x = - 1

\frac{1}{x ^{2} - 1} (x^{2} + x)

e^{x} + x^{2}

f (x) = 5 I n (4 x)

f (x) = 6 x^{2} + 6 x - 12

f (x) = x + 2

What is the critical f(x)= 1/(x+3)-1/((x+3)^2) ?
The critical f(x)= 1/(x+3)-1/((x+3)^2) is x=-1