What is the range of f(x)= 7/(3+e^x) ?

The range of f(x)= 7/(3+e^x) is 0<f(x)< 7/3

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range of f(x)= 7/(3+e^x)

range

f (x) = \frac{7}{3 + e ^{x}}

0 < f (x) < \frac{7}{3}

Interval Notation

(0, \frac{7}{3})

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{7}{3 + e ^{x}}

Inverse of

\frac{7}{3 + e ^{x}} : ln (\frac{7}{x} - 3)

ln (\frac{7}{x} - 3)

Find the domain of each inverse function

Domain of

ln (\frac{7}{x} - 3) : 0 < x < \frac{7}{3}

Combine the intervals

0 < f (x) < \frac{7}{3}

f (x) = 4 (x - 3) - 1

y = \frac{4}{7} x + 5, (4, - 7)

f (x) = x^{2} + 2 x + 7

y = arcsin (x)

f (x) = - (x + 2)^{2} + 6

What is the range of f(x)= 7/(3+e^x) ?
The range of f(x)= 7/(3+e^x) is 0<f(x)< 7/3