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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

(8 - x) (x + 1)^{2}

f (x) = x^{3} + 11 x + 10

f (x) = 3 sin (x), 0 \leq x \leq 2 π

\frac{e ^{x}}{8 + e ^{x}}

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)