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

The extreme f(x)= x/(x^2-x+2) is Minimum(-sqrt(2),-(2sqrt(2)-1)/7),Maximum(sqrt(2),(2sqrt(2)+1)/7)

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

extreme points

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

Minimum (- 2, - \frac{2 2 - 1}{7}), Maximum (2, \frac{2 2 + 1}{7})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x}{x ^{2} - x + 2} : - \frac{2 2 - 1}{7}

Minimum (- 2, - \frac{2 2 - 1}{7})

Plug

x = 2

into

\frac{x}{x ^{2} - x + 2} : \frac{2 2 + 1}{7}

Maximum (2, \frac{2 2 + 1}{7})

Minimum (- 2, - \frac{2 2 - 1}{7}), Maximum (2, \frac{2 2 + 1}{7})

f (x) = 7 x - 28 x

f (x) = ((x y) - λ (6 x + y - 20))

y = \frac{x + 3}{2}

f (x, y) = 3 x^{2} - x y + y^{2} - 11 x

f (x, y) = (x - 1)^{2} - (y - 1)^{2}

What is the extreme f(x)= x/(x^2-x+2) ?
The extreme f(x)= x/(x^2-x+2) is Minimum(-sqrt(2),-(2sqrt(2)-1)/7),Maximum(sqrt(2),(2sqrt(2)+1)/7)