What is the inflection f(x)=ln(x)+2x^2 ?

The inflection f(x)=ln(x)+2x^2 is (1/2 ,-ln(2)+1/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inflection f(x)=ln(x)+2x^2

inflection points

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

(\frac{1}{2}, - ln (2) + \frac{1}{2})

Solution steps

f^{''} (x) = - \frac{1}{x ^{2}} + 4

Find intervals:

Concave Downward

: 0 < x < \frac{1}{2},

Concave Upward

: \frac{1}{2} < x < \infty

Plug

x = \frac{1}{2}

into

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

(\frac{1}{2}, - ln (2) + \frac{1}{2})

f (x) = \frac{2}{2 + x}

f (x) = 1 0^{x}

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

(1) (5.4)

f (x) = 5 x^{2}

What is the inflection f(x)=ln(x)+2x^2 ?
The inflection f(x)=ln(x)+2x^2 is (1/2 ,-ln(2)+1/2)