What is the extreme f(x)=(5x)/(x^2+16) ?

The extreme f(x)=(5x)/(x^2+16) is Minimum(-4,-5/8),Maximum(4, 5/8)

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extreme f(x)=(5x)/(x^2+16)

extreme points

f (x) = \frac{5 x}{x ^{2} + 16}

Minimum (- 4, - \frac{5}{8}), Maximum (4, \frac{5}{8})

Solution steps

f^{'} (x) = \frac{5 ( - x ^{2} + 16 )}{( x ^{2} + 16 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = - 4

into

\frac{5 x}{x ^{2} + 16} : - \frac{5}{8}

Minimum (- 4, - \frac{5}{8})

Plug

x = 4

into

\frac{5 x}{x ^{2} + 16} : \frac{5}{8}

Maximum (4, \frac{5}{8})

Minimum (- 4, - \frac{5}{8}), Maximum (4, \frac{5}{8})

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

f (x) = 3 x^{2} ln (\frac{x}{2})

f (x, y) = \frac{1}{2} x^{2} + 2 y^{2} - 4 x - 16 y

f (x) = \frac{x + 5}{x ^{2} - x - 30}

f (x) = - x^{2} + 9, - 4 \leq x \leq 3

What is the extreme f(x)=(5x)/(x^2+16) ?
The extreme f(x)=(5x)/(x^2+16) is Minimum(-4,-5/8),Maximum(4, 5/8)