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

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

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

range

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

Inverse of

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

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

Find the domain of each inverse function

Domain of

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

Combine the intervals

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

f (x) = x - 1

(- \frac{1}{3}, \frac{2}{3}), (- \frac{11}{3}, - \frac{8}{3})

(3, 5), (- 1, 5)

f (x) = \frac{2 x + 5}{x + 2}

f (x) = \frac{1}{arccos ( t - 2 )}

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