What is the extreme f(x)= x/(x^2+49) ?

The extreme f(x)= x/(x^2+49) is Minimum(-7,-1/14),Maximum(7, 1/14)

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extreme f(x)= x/(x^2+49)

extreme points

f (x) = \frac{x}{x ^{2} + 49}

Minimum (- 7, - \frac{1}{14}), Maximum (7, \frac{1}{14})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 49}{( x ^{2} + 49 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 7,

Increasing

: - 7 < x < 7,

Decreasing

: 7 < x < \infty

Plug

x = - 7

into

\frac{x}{x ^{2} + 49} : - \frac{1}{14}

Minimum (- 7, - \frac{1}{14})

Plug

x = 7

into

\frac{x}{x ^{2} + 49} : \frac{1}{14}

Maximum (7, \frac{1}{14})

Minimum (- 7, - \frac{1}{14}), Maximum (7, \frac{1}{14})

What is the extreme f(x)= x/(x^2+49) ?
The extreme f(x)= x/(x^2+49) is Minimum(-7,-1/14),Maximum(7, 1/14)