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

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

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

extreme points

f (x) = \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})

z (b, v) = b^{2 v}

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

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

f (x) = 9 x^{2} - 4

x^{5} - 5 x^{4}

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