What is the extreme x/(x^2+16),0<= x<= 8 ?

The extreme x/(x^2+16),0<= x<= 8 is Minimum(0,0),Maximum(4, 1/8),Minimum(8, 1/10)

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extreme x/(x^2+16),0<= x<= 8

extreme

Minimum (0, 0), Maximum (4, \frac{1}{8}), Minimum (8, \frac{1}{10})

Solution steps

Find the critical points:

x = 0, x = 4, x = 8

Domain of

\frac{x}{x ^{2} + 16}, 0 \leq x \leq 8 : 0 \leq x \leq 8

Find intervals:

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 8

Plug the extreme point

x = 0

into

\frac{x}{x ^{2} + 16} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 4

into

\frac{x}{x ^{2} + 16} \Rightarrow y = \frac{1}{8}

Maximum (4, \frac{1}{8})

Plug the extreme point

x = 8

into

\frac{x}{x ^{2} + 16} \Rightarrow y = \frac{1}{10}

Minimum (8, \frac{1}{10})

Minimum (0, 0), Maximum (4, \frac{1}{8}), Minimum (8, \frac{1}{10})

What is the extreme x/(x^2+16),0<= x<= 8 ?
The extreme x/(x^2+16),0<= x<= 8 is Minimum(0,0),Maximum(4, 1/8),Minimum(8, 1/10)