What is the extreme x+ln(x^2+1) ?

The extreme x+ln(x^2+1) is Saddle(-1,-1+ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme x+ln(x^2+1)

extreme points

x + ln (x^{2} + 1)

Saddle (- 1, - 1 + ln (2))

Solution steps

Find the critical points:

x = - 1

Domain of

x + ln (x^{2} + 1) : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - 1,

Increasing

: - 1 < x < \infty

Plug the extreme point

x = - 1

into

x + ln (x^{2} + 1) \Rightarrow y = - 1 + ln (2)

Saddle (- 1, - 1 + ln (2))

V (B, h) = \frac{B h}{3}

x^{3} + 3 x^{2} + 2 x

f (x) = (\frac{1}{x}) ln (x)

f (x) = - 8 sin (4 x)

f (x) = 16 - 4 x - \frac{4}{x}, [0, \infty]

What is the extreme x+ln(x^2+1) ?
The extreme x+ln(x^2+1) is Saddle(-1,-1+ln(2))