What is the extreme f(x)=(6x)/(x^2+9) ?

The extreme f(x)=(6x)/(x^2+9) is Minimum(-3,-1),Maximum(3,1)

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extreme f(x)=(6x)/(x^2+9)

extreme points

f (x) = \frac{6 x}{x ^{2} + 9}

Minimum (- 3, - 1), Maximum (3, 1)

Solution steps

f^{'} (x) = \frac{6 ( - x ^{2} + 9 )}{( x ^{2} + 9 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{6 x}{x ^{2} + 9} : - 1

Minimum (- 3, - 1)

Plug

x = 3

into

\frac{6 x}{x ^{2} + 9} : 1

Maximum (3, 1)

Minimum (- 3, - 1), Maximum (3, 1)

What is the extreme f(x)=(6x)/(x^2+9) ?
The extreme f(x)=(6x)/(x^2+9) is Minimum(-3,-1),Maximum(3,1)