What is the extreme f(x)=((x^2))/(x-1) ?

The extreme f(x)=((x^2))/(x-1) is Maximum(0,0),Minimum(2,4)

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extreme f(x)=((x^2))/(x-1)

extreme points

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

Maximum (0, 0), Minimum (2, 4)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x - 1} : 0

Maximum (0, 0)

Plug

x = 2

into

\frac{x ^{2}}{x - 1} : 4

Minimum (2, 4)

Maximum (0, 0), Minimum (2, 4)

What is the extreme f(x)=((x^2))/(x-1) ?
The extreme f(x)=((x^2))/(x-1) is Maximum(0,0),Minimum(2,4)