What is the extreme f(x)=8x-8ln(x^2),(0,+infinity) ?

The extreme f(x)=8x-8ln(x^2),(0,+infinity) is Minimum(2,16-16ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=8x-8ln(x^2),(0,+infinity)

extreme points

f (x) = 8 x - 8 ln (x^{2}), (0, + \infty)

Minimum (2, 16 - 16 ln (2))

Solution steps

Find the critical points:

x = 2

Domain of

8 x - 8 ln (x^{2}), (0, + \infty) : x > 0

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 2

into

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

Minimum (2, 16 - 16 ln (2))

f (x, y) = 7 x^{2} y - 8 x y^{3} + 9 y^{3}

\frac{8 - x ^{3}}{x ^{2}}

f (x) = x^{2} lo g_{4} (x)

f (x) = \frac{2 - x ^{2}}{2 x + 1} + 7

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

What is the extreme f(x)=8x-8ln(x^2),(0,+infinity) ?
The extreme f(x)=8x-8ln(x^2),(0,+infinity) is Minimum(2,16-16ln(2))