What is the extreme f(x)=(12x)/(9+x^2) ?

The extreme f(x)=(12x)/(9+x^2) is Minimum(-3,-2),Maximum(3,2)

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extreme f(x)=(12x)/(9+x^2)

extreme points

f (x) = \frac{12 x}{9 + x ^{2}}

Minimum (- 3, - 2), Maximum (3, 2)

Solution steps

f^{'} (x) = \frac{12 ( 9 - x ^{2} )}{( 9 + x ^{2} ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{12 x}{9 + x ^{2}} : - 2

Minimum (- 3, - 2)

Plug

x = 3

into

\frac{12 x}{9 + x ^{2}} : 2

Maximum (3, 2)

Minimum (- 3, - 2), Maximum (3, 2)

f (x, y) = 4 - 2 x + 4 y - x^{2} - 4 y^{2}

f (x) = 2 x^{\frac{5}{3}} - 5 x^{\frac{4}{3}}

f (x) = 2 (3 x + 2)^{2} - 3 x^{4}

y = x (x + 4)^{3}

f (x) = \frac{4 x}{x ^{2} - 49}

What is the extreme f(x)=(12x)/(9+x^2) ?
The extreme f(x)=(12x)/(9+x^2) is Minimum(-3,-2),Maximum(3,2)