What is the extreme f(x)=(14x)/(x^2+49) ?

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

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extreme f(x)=(14x)/(x^2+49)

extreme points

f (x) = \frac{14 x}{x ^{2} + 49}

Minimum (- 7, - 1), Maximum (7, 1)

Solution steps

f^{'} (x) = \frac{14 ( - 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{14 x}{x ^{2} + 49} : - 1

Minimum (- 7, - 1)

Plug

x = 7

into

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

Maximum (7, 1)

Minimum (- 7, - 1), Maximum (7, 1)

\frac{x ^{2} - 2 x - 1}{x + 1}

(- 1, 3), (8, - 5)

- 8 x^{2} + 4 x - 2

\frac{x ^{2} - 9 x + 39}{x - 7}

f (x) = \frac{1}{x + 6}

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