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

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

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

extreme points

f (x) = \frac{x}{x ^{2} + 16}, 0 \leq x \leq 8

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})

f (x, y) = x^{2} + x y + \frac{1}{2} y^{2} - 2 x + y

f (x) = \frac{x ^{2} + 11}{x + 5}

f (x) = 3 - x [- 1.2]

f (x) = 4 - x

f (x, y) = (x - 2)^{2} + (y + 1)^{2}

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