What is the extreme (4x)/(1+9x^2) ?

The extreme (4x)/(1+9x^2) is Minimum(-1/3 ,-2/3),Maximum(1/3 , 2/3)

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extreme (4x)/(1+9x^2)

extreme points

\frac{4 x}{1 + 9 x ^{2}}

Minimum (- \frac{1}{3}, - \frac{2}{3}), Maximum (\frac{1}{3}, \frac{2}{3})

Solution steps

f^{'} (x) = \frac{4 ( 1 - 9 x ^{2} )}{( 1 + 9 x ^{2} ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{3},

Increasing

: - \frac{1}{3} < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < \infty

Plug

x = - \frac{1}{3}

into

\frac{4 x}{1 + 9 x ^{2}} : - \frac{2}{3}

Minimum (- \frac{1}{3}, - \frac{2}{3})

Plug

x = \frac{1}{3}

into

\frac{4 x}{1 + 9 x ^{2}} : \frac{2}{3}

Maximum (\frac{1}{3}, \frac{2}{3})

Minimum (- \frac{1}{3}, - \frac{2}{3}), Maximum (\frac{1}{3}, \frac{2}{3})

f (x, y) = y x - y^{2} - x + 6 y

f (x) = \frac{1}{x}, (- 2, 3)

y = x - ln (x)

f (x, y) = y^{3} - x^{2} + 2 x y - 5 y + 5

y = \frac{x ^{3} + 4}{x ^{2}}

What is the extreme (4x)/(1+9x^2) ?
The extreme (4x)/(1+9x^2) is Minimum(-1/3 ,-2/3),Maximum(1/3 , 2/3)