What is the critical f(x)=(((x-6)^{(1/3)}))/(x-2) ?

The critical f(x)=(((x-6)^{(1/3)}))/(x-2) is x=6,x=8

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critical f(x)=(((x-6)^{(1/3)}))/(x-2)

critical points

f (x) = \frac{( ( x - 6 ) ^{(\frac{1}{3})} )}{x - 2}

x = 6, x = 8

Solution steps

\frac{( x - 6 ) ^{\frac{1}{3}}}{x - 2}

Find where

f^{'} (x)

is equal to zero or undefined

x = 8, x = 2, x = 6

Identify critical points not in

f (x)

domain

Domain of

\frac{( x - 6 ) ^{\frac{1}{3}}}{x - 2} : x < 2 or x > 2

\frac{( x - 6 ) ^{\frac{1}{3}}}{x - 2}

is not defined at

x = 2

, therefore:

x = 6, x = 8

y = - x^{4} - x^{3} + x^{2} + 3 x + 2

f (x) = x^{3} - 5 x^{2} + 3 x + 10

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

f (x) = x^{3} + 3 x^{2} - 45 x

y = x ln (x)

What is the critical f(x)=(((x-6)^{(1/3)}))/(x-2) ?
The critical f(x)=(((x-6)^{(1/3)}))/(x-2) is x=6,x=8