What is the range of f(x)= 1/(5+e^{3x)} ?

The range of f(x)= 1/(5+e^{3x)} is 0<f(x)< 1/5

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range of f(x)= 1/(5+e^{3x)}

range

f (x) = \frac{1}{5 + e ^{3 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 ^{3 x}}

Inverse of

\frac{1}{5 + e ^{3 x}} : \frac{ln ( \frac{1}{x} - 5 )}{3}

\frac{ln ( \frac{1}{x} - 5 )}{3}

Find the domain of each inverse function

Domain of

\frac{ln ( \frac{1}{x} - 5 )}{3} : 0 < x < \frac{1}{5}

Combine the intervals

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

\frac{6 x + 7}{5 x - 6}

X + ln (X^{2} - 1)

y = lo g_{3} (4 x + 18) - 3

m = - 4, (7, 8)

y = \frac{( x + 1 )}{2 x + 3}

What is the range of f(x)= 1/(5+e^{3x)} ?
The range of f(x)= 1/(5+e^{3x)} is 0<f(x)< 1/5