What is the critical f(x)=(x^2)/(3-x) ?

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

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critical f(x)=(x^2)/(3-x)

critical points

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

x = 0, x = 6

Solution steps

\frac{x ^{2}}{3 - x}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 6, x = 3

Identify critical points not in

f (x)

domain

Domain of

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

\frac{x ^{2}}{3 - x}

is not defined at

x = 3

, therefore:

x = 0, x = 6

y = \frac{ln ( x )}{x ^{2}}

f (x) = x^{3} - 6 x y + 8 y^{3}

f (x) = 3 x

f (x, y) = x^{3} - 6 x y + 8 y^{3}

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

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