What is the critical f(x)=(x^2+8x-4)/(x-2) ?

The critical f(x)=(x^2+8x-4)/(x-2) is x=-2,x=6

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critical f(x)=(x^2+8x-4)/(x-2)

critical points

f (x) = \frac{x ^{2} + 8 x - 4}{x - 2}

x = - 2, x = 6

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 2, x = 6, x = 2

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} + 8 x - 4}{x - 2} : x < 2 or x > 2

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

is not defined at

x = 2

, therefore:

x = - 2, x = 6

\frac{x - 3}{x + 3}

f (x) = 6 x^{2} + 7 x - 5

f (x) = lo g_{3} (x - 1) + 0.239784

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

- x^{2} + 3

What is the critical f(x)=(x^2+8x-4)/(x-2) ?
The critical f(x)=(x^2+8x-4)/(x-2) is x=-2,x=6