What is the extreme f(x)=(3x)/(x^2+1),0<= x<= 3 ?

The extreme f(x)=(3x)/(x^2+1),0<= x<= 3 is Minimum(0,0),Maximum(1, 3/2),Minimum(3, 9/10)

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extreme f(x)=(3x)/(x^2+1),0<= x<= 3

extreme points

f (x) = \frac{3 x}{x ^{2} + 1}, 0 \leq x \leq 3

Minimum (0, 0), Maximum (1, \frac{3}{2}), Minimum (3, \frac{9}{10})

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

\frac{3 x}{x ^{2} + 1}, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 3

Plug the extreme point

x = 0

into

\frac{3 x}{x ^{2} + 1} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

\frac{3 x}{x ^{2} + 1} \Rightarrow y = \frac{3}{2}

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

Plug the extreme point

x = 3

into

\frac{3 x}{x ^{2} + 1} \Rightarrow y = \frac{9}{10}

Minimum (3, \frac{9}{10})

Minimum (0, 0), Maximum (1, \frac{3}{2}), Minimum (3, \frac{9}{10})

f (x) = 6 cos (x)

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

2 x^{3} + 3 x^{2} - 12 x - 18

f (x) = 4.5 x^{2} - 7.1 x + 3.9

y = 6 - 2 x^{2} - z^{2}

What is the extreme f(x)=(3x)/(x^2+1),0<= x<= 3 ?
The extreme f(x)=(3x)/(x^2+1),0<= x<= 3 is Minimum(0,0),Maximum(1, 3/2),Minimum(3, 9/10)