What is the extreme f(x)=(8x)/(x^2+1) ?

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

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extreme f(x)=(8x)/(x^2+1)

extreme

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{8 x}{x ^{2} + 1} : - 4

Minimum (- 1, - 4)

Plug

x = 1

into

\frac{8 x}{x ^{2} + 1} : 4

Maximum (1, 4)

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

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