en

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^2+1))

Solution

extreme points

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

Solution

Minimum (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

ln (x^{2} + 1) : 0

Minimum (0, 0)

Popular Examples

extreme f(x)=3x+5x^{-1}extreme points

f (x) = 3 x + 5 x^{- 1}

extreme f(x)=(x-1)^5 extreme points

f (x) = (x - 1)^{5}

extreme f(x,y)=3xy-2x^3y^3 extreme points

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

extreme f(x)=x^2=-4ay extreme points

f (x) = x^{2} = - 4 a y

extreme-x^2-4y^2+8x-8y-16 extreme points

- x^{2} - 4 y^{2} + 8 x - 8 y - 16

Frequently Asked Questions (FAQ)

What is the extreme f(x)=(ln(x^2+1)) ?
The extreme f(x)=(ln(x^2+1)) is Minimum(0,0)