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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, - 1)

Plug

x = 2

into

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

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

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

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

f (x) = ln (9 - x^{2} - 9 y^{2})

x^{4} + y^{4} - 4 x y + 5

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

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

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