What is the extreme y= x/(x^2-x+4) ?

The extreme y= x/(x^2-x+4) is Minimum(-2,-1/5),Maximum(2, 1/3)

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extreme y= x/(x^2-x+4)

extreme points

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

Minimum (- 2, - \frac{1}{5}), Maximum (2, \frac{1}{3})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 4}{( x ^{2} - x + 4 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x}{x ^{2} - x + 4} : - \frac{1}{5}

Minimum (- 2, - \frac{1}{5})

Plug

x = 2

into

\frac{x}{x ^{2} - x + 4} : \frac{1}{3}

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

Minimum (- 2, - \frac{1}{5}), Maximum (2, \frac{1}{3})

\frac{4 x}{1 + 9 x ^{2}}

f (x, y) = y x - y^{2} - x + 6 y

f (x) = \frac{1}{x}, (- 2, 3)

y = x - ln (x)

f (x, y) = y^{3} - x^{2} + 2 x y - 5 y + 5

What is the extreme y= x/(x^2-x+4) ?
The extreme y= x/(x^2-x+4) is Minimum(-2,-1/5),Maximum(2, 1/3)