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

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

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

extreme points

y = \frac{x}{x ^{2} - x + 1}

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, - \frac{1}{3})

Plug

x = 1

into

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

Maximum (1, 1)

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

f (x) = 4 x^{3} - 12 x^{2} + 8 x

f (x, y) = \frac{16 - x ^{2} - 4 y ^{2}}{2}

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

y = - x^{2} + 4 x + 2

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

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