What is the critical f(x)=(x^2)/(x-5) ?

The critical f(x)=(x^2)/(x-5) is x=0,x=10

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critical f(x)=(x^2)/(x-5)

critical points

f (x) = \frac{x ^{2}}{x - 5}

x = 0, x = 10

Solution steps

\frac{x ^{2}}{x - 5}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 10, x = 5

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{x - 5} : x < 5 or x > 5

\frac{x ^{2}}{x - 5}

is not defined at

x = 5

, therefore:

x = 0, x = 10

f (x) = 4 x e^{3 x}

x^{4} e^{- 2 x}

\frac{1}{3} x^{3} - x^{2} - 3 x + 4

\frac{x ^{2} + x - 38}{x ^{2} - 25}

y = 2 x - x^{2}

What is the critical f(x)=(x^2)/(x-5) ?
The critical f(x)=(x^2)/(x-5) is x=0,x=10