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

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

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

extreme points

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

Maximum (0, - 2), Minimum (2, 2)

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} - 2 x + 2}{x - 1} : - 2

Maximum (0, - 2)

Plug

x = 2

into

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

Minimum (2, 2)

Maximum (0, - 2), Minimum (2, 2)

f (x) = 2 x^{3} - 9 x^{2} + 27

f (x, y) = x^{2} - x^{2} y^{2} + y^{2}

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

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

f (x, y) = 3 x^{2} + 2 y^{2}

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