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

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)

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)