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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = \frac{9 ( - 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{9 x}{x ^{2} + 9} : - \frac{3}{2}

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

Plug

x = 3

into

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

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

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

f (x) = - 6 x^{2} + 30 x - 36

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

f (x, y) = - 8 x y + 2 x^{4} + 2 y^{4}

f (x) = 2 x - sin (x), 0 \leq x \leq π

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 8

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