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

The extreme f(x)=(x^2-x+1)/(x-1) is Maximum(0,-1),Minimum(2,3)

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

extreme points

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

Maximum (0, - 1), Minimum (2, 3)

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}{x - 1} : - 1

Maximum (0, - 1)

Plug

x = 2

into

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

Minimum (2, 3)

Maximum (0, - 1), Minimum (2, 3)

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

\frac{d}{d x} (5 x^{2} y^{4} - 23 x y^{3} + 4 y^{5})

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

f (x) = x^{3} - x^{2}, 0 \leq x \leq 5

f (x, y) = 72 x^{4} + y^{2} - 24 x y

What is the extreme f(x)=(x^2-x+1)/(x-1) ?
The extreme f(x)=(x^2-x+1)/(x-1) is Maximum(0,-1),Minimum(2,3)