What is the extreme f(x)=(8x)/(x^2+16) ?

The extreme f(x)=(8x)/(x^2+16) is Minimum(-4,-1),Maximum(4,1)

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extreme f(x)=(8x)/(x^2+16)

extreme points

f (x) = \frac{8 x}{x ^{2} + 16}

Minimum (- 4, - 1), Maximum (4, 1)

Solution steps

f^{'} (x) = \frac{8 ( - 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{8 x}{x ^{2} + 16} : - 1

Minimum (- 4, - 1)

Plug

x = 4

into

\frac{8 x}{x ^{2} + 16} : 1

Maximum (4, 1)

Minimum (- 4, - 1), Maximum (4, 1)

\frac{( x ^{2} - 1 )}{( x - 3 ) ^{2}}

s (t) = - 4.9 t^{2} + 29 t + 2

y = (x + 3)^{2} - 7

f (x) = 3 x - \frac{1}{2} x^{2} - \frac{1}{2} ln (x)

f (x) = 20.85 x y + 10.7 x^{2} + 51.75 y^{2}

What is the extreme f(x)=(8x)/(x^2+16) ?
The extreme f(x)=(8x)/(x^2+16) is Minimum(-4,-1),Maximum(4,1)