What is the critical (x^2-x-2)/(x^2-6x+9) ?

The critical (x^2-x-2)/(x^2-6x+9) is x= 7/5

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critical (x^2-x-2)/(x^2-6x+9)

critical points

\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9}

x = \frac{7}{5}

Solution steps

\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{7}{5}, x = 3

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9} : x < 3 or x > 3

\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9}

is not defined at

x = 3

, therefore:

x = \frac{7}{5}

(- 3, 4), (- 6, - 1)

at (- 7 - 5), y = 5

f (x) = x^{2} + 4 x + 6

5 x^{2} - 5

2 x^{2} + 4 x - 1

What is the critical (x^2-x-2)/(x^2-6x+9) ?
The critical (x^2-x-2)/(x^2-6x+9) is x= 7/5