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

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

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

critical points

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

x = \frac{3}{2}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{3}{2}, x = 1, x = 2

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = 1, x = 2

, therefore:

x = \frac{3}{2}

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

f (x) = x^{3} - 6 x^{2} + 9 x + 15

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

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

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