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

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

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

critical points

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

x = - 3, x = - 1

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = - 1, x = - 2

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = - 2

, therefore:

x = - 3, x = - 1

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

(- 4, - 1), (- 2, 5)

y = - 2 x + 3

f (x) = \frac{1}{2} x^{2} + x - \frac{7}{2}

3 cos (\frac{1}{3} x)

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