What is the extreme f(x)=(8x)/(x^2+9),0<= x<= 9 ?

The extreme f(x)=(8x)/(x^2+9),0<= x<= 9 is Minimum(0,0),Maximum(3, 4/3),Minimum(9, 4/5)

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extreme f(x)=(8x)/(x^2+9),0<= x<= 9

extreme points

f (x) = \frac{8 x}{x ^{2} + 9}, 0 \leq x \leq 9

Minimum (0, 0), Maximum (3, \frac{4}{3}), Minimum (9, \frac{4}{5})

Solution steps

Find the critical points:

x = 0, x = 3, x = 9

Domain of

\frac{8 x}{x ^{2} + 9}, 0 \leq x \leq 9 : 0 \leq x \leq 9

Find intervals:

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < 9

Plug the extreme point

x = 0

into

\frac{8 x}{x ^{2} + 9} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 3

into

\frac{8 x}{x ^{2} + 9} \Rightarrow y = \frac{4}{3}

Maximum (3, \frac{4}{3})

Plug the extreme point

x = 9

into

\frac{8 x}{x ^{2} + 9} \Rightarrow y = \frac{4}{5}

Minimum (9, \frac{4}{5})

Minimum (0, 0), Maximum (3, \frac{4}{3}), Minimum (9, \frac{4}{5})

f (x) = - 3 x^{2} - 2 y^{2} + 3 x - 4 y + 5

f (x) = x^{3} - 27 x + 55

\frac{1}{2} x^{\frac{2}{3}} (2 x - 5)

f (x) = x + \frac{25}{x} + 4, 1 \leq x \leq 50

f (x, y) = 400 - 4 x^{2} - 81 y^{2}

What is the extreme f(x)=(8x)/(x^2+9),0<= x<= 9 ?
The extreme f(x)=(8x)/(x^2+9),0<= x<= 9 is Minimum(0,0),Maximum(3, 4/3),Minimum(9, 4/5)