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

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

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critical y=(x^2+2)/(2x-1)

critical points

y = \frac{x ^{2} + 2}{2 x - 1}

x = - 1, x = 2

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 2, x = \frac{1}{2}

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = \frac{1}{2}

, therefore:

x = - 1, x = 2

f (x) = x^{3} + 9 x^{2} + b

x^{\frac{8}{3}} - 8 x^{\frac{1}{3}}

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

e^{- x} x^{2} + e^{- x}

f (x) = 3 x + \frac{2}{x} + 6

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