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

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

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

critical points

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

x = - 1, x = 5

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 5, x = 2

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = 2

, therefore:

x = - 1, x = 5

f (x) = \frac{3 - 4 x}{x + 1}

(6 x - 4)^{\frac{1}{2}}

\frac{x ^{2} + 4 x + 8}{4 x}

f (x) = \frac{1 - 3 t}{6 + t}

f (x) = (lo g_{8} (x)) - 8

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