What is the extreme f(x)=(ln(x))/(8x),1<= x<= 4 ?

The extreme f(x)=(ln(x))/(8x),1<= x<= 4 is Minimum(1,0),Maximum(e, 1/(8e)),Minimum(4,(ln(2))/(16))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(ln(x))/(8x),1<= x<= 4

extreme points

f (x) = \frac{ln ( x )}{8 x}, 1 \leq x \leq 4

Minimum (1, 0), Maximum (e, \frac{1}{8 e}), Minimum (4, \frac{ln ( 2 )}{16})

Solution steps

Find the critical points:

x = 1, x = e, x = 4

Domain of

\frac{ln ( x )}{8 x}, 1 \leq x \leq 4 : 1 \leq x \leq 4

Find intervals:

Increasing

: 1 < x < e,

Decreasing

: e < x < 4

Plug the extreme point

x = 1

into

\frac{ln ( x )}{8 x} \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = e

into

\frac{ln ( x )}{8 x} \Rightarrow y = \frac{1}{8 e}

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

Plug the extreme point

x = 4

into

\frac{ln ( x )}{8 x} \Rightarrow y = \frac{ln ( 2 )}{16}

Minimum (4, \frac{ln ( 2 )}{16})

Minimum (1, 0), Maximum (e, \frac{1}{8 e}), Minimum (4, \frac{ln ( 2 )}{16})

f (x) = \frac{1 + 3 x}{4 + 5 x ^{2}}

f (x) = x^{2} + y^{2} - 6 x + 3 y

f (x) = (2 x^{2} + 1)^{2}

f (x, y) = \frac{100}{1 + x ^{2} + y ^{2}}

f (x) = \frac{1}{6} x^{3} - x^{2} + 6 x + 11

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