What is the extreme f(x)=(7x)/(x^2+9),0<= x<= 9 ?

The extreme f(x)=(7x)/(x^2+9),0<= x<= 9 is Minimum(0,0),Maximum(3, 7/6),Minimum(9, 7/10)

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extreme f(x)=(7x)/(x^2+9),0<= x<= 9

extreme points

f (x) = \frac{7 x}{x ^{2} + 9}, 0 \leq x \leq 9

Minimum (0, 0), Maximum (3, \frac{7}{6}), Minimum (9, \frac{7}{10})

Solution steps

Find the critical points:

x = 0, x = 3, x = 9

Domain of

\frac{7 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{7 x}{x ^{2} + 9} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 3

into

\frac{7 x}{x ^{2} + 9} \Rightarrow y = \frac{7}{6}

Maximum (3, \frac{7}{6})

Plug the extreme point

x = 9

into

\frac{7 x}{x ^{2} + 9} \Rightarrow y = \frac{7}{10}

Minimum (9, \frac{7}{10})

Minimum (0, 0), Maximum (3, \frac{7}{6}), Minimum (9, \frac{7}{10})

2 x^{3} + 4 x^{2} - 1

f (x) = x \cdot (ln (x))^{5}

f (x, y) = 40000 x + 30000 y - 8 x^{2} - 15 y^{2} - 10 x y

f (x) = e^{x^{3} - 3 x}

f (x) = x + \frac{30}{x}

What is the extreme f(x)=(7x)/(x^2+9),0<= x<= 9 ?
The extreme f(x)=(7x)/(x^2+9),0<= x<= 9 is Minimum(0,0),Maximum(3, 7/6),Minimum(9, 7/10)