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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < y < 0,

Increasing

: 0 < y < 2,

Decreasing

: 2 < y < \infty

Plug

y = 0

into

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

Minimum (0, - 1)

Plug

y = 2

into

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

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

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

F (x, y) = 15 x^{2} - 22 x y + 24 x + 8 y^{2} - 16 y

x^{6}

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

f (x) = x^{2} - 8 x + 40

f (x) = x^{2} - 8 x + 18

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