What is the range of f(x)= 1/(3+e^{2x)} ?

The range of f(x)= 1/(3+e^{2x)} is 0<f(x)< 1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

range of f(x)= 1/(3+e^{2x)}

range

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

0 < f (x) < \frac{1}{3}

Interval Notation

(0, \frac{1}{3})

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{1}{3 + e ^{2 x}}

Inverse of

\frac{1}{3 + e ^{2 x}} : \frac{ln ( \frac{1}{x} - 3 )}{2}

\frac{ln ( \frac{1}{x} - 3 )}{2}

Find the domain of each inverse function

Domain of

\frac{ln ( \frac{1}{x} - 3 )}{2} : 0 < x < \frac{1}{3}

Combine the intervals

0 < f (x) < \frac{1}{3}

What is the range of f(x)= 1/(3+e^{2x)} ?
The range of f(x)= 1/(3+e^{2x)} is 0<f(x)< 1/3