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 points

y = \frac{x}{x ^{2} + 16}

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})

- x^{3} (x - 3)

f (x) = \frac{4}{3} x^{3} - \frac{13}{2} x^{2} - 3 x + 11

f (x) = 3 x^{3} - 14 x + 20 x - 8

f (x) = (2 x^{2} - 5 x)^{3}

f (x) = x^{\frac{4}{5}} (x - 2)

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)