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, y) = 2 x^{2} + x \cdot y^{3} - 4 y

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

f (y) = 2 x^{2} - 9 x + 5

f (x y) = 2 x^{2} + \frac{3}{2} y^{2} + 3 x y - 10 x - 9 y + 2

2 x + 1 - 2 x

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)