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

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

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

extreme points

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

Minimum (- 2, - 1), Maximum (2, 1)

Solution steps

f^{'} (x) = \frac{4 ( - 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{4 x}{x ^{2} + 4} : - 1

Minimum (- 2, - 1)

Plug

x = 2

into

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

Maximum (2, 1)

Minimum (- 2, - 1), Maximum (2, 1)

- 2 sin (4 x - π)

9 - x^{2}

\frac{1}{x ^{2} - 9}

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

\frac{1}{7 x ^{3} - 5 x}

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