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

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

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extreme y= x/(x^2+16)

extreme

Minimum (- 4, - \frac{1}{8}), Maximum (4, \frac{1}{8})

Solution steps

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

Minimum (- 4, - \frac{1}{8})

Plug

x = 4

into

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

Maximum (4, \frac{1}{8})

Minimum (- 4, - \frac{1}{8}), Maximum (4, \frac{1}{8})

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