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

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

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

extreme points

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

Minimum (0, \frac{1}{2}), Maximum (2, \frac{3}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, \frac{1}{2})

Plug

x = 2

into

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

Maximum (2, \frac{3}{2})

Minimum (0, \frac{1}{2}), Maximum (2, \frac{3}{2})

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

\frac{x}{ln ( x )}, 2 \leq x \leq 10

f (x) = 10000 e^{0.17 (x - 1.62)^{2}}

f (x, y) = \frac{5 x}{1 + x ^{2} + y ^{2}}

f (x) = x^{3} - 4 x + 6

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