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

f (x) = \frac{x}{x ^{2} - 49}

y = - 4 cos (2 x + \frac{π}{3})

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

f (x) = \frac{- 4 x - 45}{9 x + 59}

5^{2 x + 1} - 2^{1 - 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