What is the extreme f(x)=(4x)/(x^2+25) ?

The extreme f(x)=(4x)/(x^2+25) is Minimum(-5,-2/5),Maximum(5, 2/5)

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extreme f(x)=(4x)/(x^2+25)

extreme points

f (x) = \frac{4 x}{x ^{2} + 25}

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

Solution steps

f^{'} (x) = \frac{4 ( - x ^{2} + 25 )}{( x ^{2} + 25 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = - 5

into

\frac{4 x}{x ^{2} + 25} : - \frac{2}{5}

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

Plug

x = 5

into

\frac{4 x}{x ^{2} + 25} : \frac{2}{5}

Maximum (5, \frac{2}{5})

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

f (x, y) = 14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

y = x^{2} - 3 x

f (x, y) = x^{2} + y^{4} + 2 x y

y = \frac{1}{3} x^{3} - 4 x - 1

f (x, y) = x^{4} + y^{4} - 4 x y + 4

What is the extreme f(x)=(4x)/(x^2+25) ?
The extreme f(x)=(4x)/(x^2+25) is Minimum(-5,-2/5),Maximum(5, 2/5)