What is the range of f(x)= 1/(5+e^{2x)} ?

The range of f(x)= 1/(5+e^{2x)} is 0<f(x)< 1/5

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range of f(x)= 1/(5+e^{2x)}

range

f (x) = \frac{1}{5 + e ^{2 x}}

0 < f (x) < \frac{1}{5}

Interval Notation

(0, \frac{1}{5})

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{1}{5 + e ^{2 x}}

Inverse of

\frac{1}{5 + e ^{2 x}} : \frac{ln ( \frac{1}{x} - 5 )}{2}

\frac{ln ( \frac{1}{x} - 5 )}{2}

Find the domain of each inverse function

Domain of

\frac{ln ( \frac{1}{x} - 5 )}{2} : 0 < x < \frac{1}{5}

Combine the intervals

0 < f (x) < \frac{1}{5}

3 x^{4} - 15

f (x) = \frac{13 x ^{2}}{7 x ^{2} + 6}

y = \frac{1}{2} x + 2

f (x) = \frac{4 x}{x ^{2} + 81}

(18, 1), (2, - 1)

What is the range of f(x)= 1/(5+e^{2x)} ?
The range of f(x)= 1/(5+e^{2x)} is 0<f(x)< 1/5