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

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

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range of 7/(3+e^x)

range

\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) = \frac{( x + 1 ) ^{2}}{x - 4}

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

f (x) = \frac{( 6 - 2 x )}{x + 3}

\frac{3 x + 4}{2 x - 3}

f (x) = 1 + \frac{1}{x}