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

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

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

critical points

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

x = - 3

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = - 3, x = 3

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = - 3, x = 3

, therefore:

x = - 3

y = 2 x^{3} - 9 x^{2} + 12 x

y = x^{2} + 1

f (x) = x^{3} (x - 5)^{2}

y = - 2 csc (3 x + π) - 1

f (x) = 6 x^{5} + 33 x^{4} + 30 x^{3} + 100

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