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

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

range

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

0 < f (x) < \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}

x^{2} - 2 x + 3

f (x) = x^{2} - 5 x

f (x) = 3 + 4 x^{2} \cdot e^{- x}

- 3 sin (2 x + \frac{π}{2})

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

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