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

The critical f(x)=(x^2)/((x-2)(x-6)) is x=0,x=3

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critical f(x)=(x^2)/((x-2)(x-6))

critical points

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

x = 0, x = 3

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 3, x = 2, x = 6

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = 2, x = 6

, therefore:

x = 0, x = 3

f (x) = I n (x - 5)

f (x) = I n (x - 3)

f (x) = (12 - 3 x) e^{x}

f (x) = 15 + 3 x^{2} - x^{4}

f (x) = 3 sin (x), - π \leq x \leq π

What is the critical f(x)=(x^2)/((x-2)(x-6)) ?
The critical f(x)=(x^2)/((x-2)(x-6)) is x=0,x=3