What is the critical f(x)= 1/(x^2-5x+6) ?

The critical f(x)= 1/(x^2-5x+6) is x= 5/2

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critical f(x)= 1/(x^2-5x+6)

critical points

f (x) = \frac{1}{x ^{2} - 5 x + 6}

x = \frac{5}{2}

Solution steps

\frac{1}{x ^{2} - 5 x + 6}

Find where

f^{'} (x)

is equal to zero or undefined

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

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{x ^{2} - 5 x + 6} : x < 2 or 2 < x < 3 or x > 3

\frac{1}{x ^{2} - 5 x + 6}

is not defined at

x = 2, x = 3

, therefore:

x = \frac{5}{2}

f (x) = 1 + 3 y

f (x) = 5 x^{2} - 3 y^{2} - 30 x + 7 y + 4 x y

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

y = \frac{1}{x - 2} - 3

f (x) = cos (x) - sin (x), (0, 2 π)

What is the critical f(x)= 1/(x^2-5x+6) ?
The critical f(x)= 1/(x^2-5x+6) is x= 5/2