What is the extreme f(x)=(7x)/(x^2+49) ?

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

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extreme f(x)=(7x)/(x^2+49)

extreme points

f (x) = \frac{7 x}{x ^{2} + 49}

Minimum (- 7, - \frac{1}{2}), Maximum (7, \frac{1}{2})

Solution steps

f^{'} (x) = \frac{7 ( - 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{7 x}{x ^{2} + 49} : - \frac{1}{2}

Minimum (- 7, - \frac{1}{2})

Plug

x = 7

into

\frac{7 x}{x ^{2} + 49} : \frac{1}{2}

Maximum (7, \frac{1}{2})

Minimum (- 7, - \frac{1}{2}), Maximum (7, \frac{1}{2})

f (x) = \frac{x - 1}{x ^{2} - 9}

f (x) = 2 x^{2} - 5 x + 1

\frac{2}{3 - x}

y = - 11 x

f (x) = x^{3} - 3 x^{2} - 9 x + 2

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