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

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

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

critical points

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

x = 0, x = 5

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 5, x = \frac{5}{2}

Identify critical points not in

f (x)

domain

Domain of

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

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

is not defined at

x = \frac{5}{2}

, therefore:

x = 0, x = 5

cr i t i c a l f (x) = 2 x ln (x)

cr i t i c a l f (x) = x 6 - x

cr i t i c a l f (x) = 3 x^{2} - 2 x + 1

cr i t i c a l x^{4} - 4 x^{3} + 4 x^{2}

cr i t i c a l t 2 - t

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