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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

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

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

Plug

x = 2

into

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

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

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

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

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

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

f (x) = ln (9 - x^{2} - 9 y^{2})

x^{4} + y^{4} - 4 x y + 5

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