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

The critical (2x-3)/(x^2-1) is x=(3-sqrt(5))/2 ,x=(3+sqrt(5))/2

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critical (2x-3)/(x^2-1)

critical points

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

x = \frac{3 - 5}{2}, x = \frac{3 + 5}{2}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{3 - 5}{2}, x = \frac{3 + 5}{2}, x = - 1, x = 1

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = - 1, x = 1

, therefore:

x = \frac{3 - 5}{2}, x = \frac{3 + 5}{2}

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

f (x) = \frac{2}{x + 13}

y = 6 x - 4

f (x) = \frac{12}{x}

f (x) = \frac{6 e ^{x}}{e ^{x} - 6}

What is the critical (2x-3)/(x^2-1) ?
The critical (2x-3)/(x^2-1) is x=(3-sqrt(5))/2 ,x=(3+sqrt(5))/2