What is the extreme f(x)= x/(x^2+64) ?

The extreme f(x)= x/(x^2+64) is Minimum(-8,-1/16),Maximum(8, 1/16)

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extreme f(x)= x/(x^2+64)

extreme points

f (x) = \frac{x}{x ^{2} + 64}

Minimum (- 8, - \frac{1}{16}), Maximum (8, \frac{1}{16})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 64}{( x ^{2} + 64 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 8,

Increasing

: - 8 < x < 8,

Decreasing

: 8 < x < \infty

Plug

x = - 8

into

\frac{x}{x ^{2} + 64} : - \frac{1}{16}

Minimum (- 8, - \frac{1}{16})

Plug

x = 8

into

\frac{x}{x ^{2} + 64} : \frac{1}{16}

Maximum (8, \frac{1}{16})

Minimum (- 8, - \frac{1}{16}), Maximum (8, \frac{1}{16})

f (x) = x + 1 - x

(x + 2)^{\frac{1}{2}}

f (x) = \frac{( \frac{x}{x + 8} )}{( \frac{x}{x + 8} ) + 8}

- 2 (x + 1)^{2} (x - 4)^{3} (x + 2)

y = \frac{x + 8}{x - 7}

What is the extreme f(x)= x/(x^2+64) ?
The extreme f(x)= x/(x^2+64) is Minimum(-8,-1/16),Maximum(8, 1/16)