What is the range of (1000)/(100+900e^{-x)} ?

The range of (1000)/(100+900e^{-x)} is 0<f(x)<10

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range of (1000)/(100+900e^{-x)}

range

\frac{1000}{100 + 900 e ^{- x}}

0 < f (x) < 10

Interval Notation

(0, 10)

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{1000}{100 + 900 e ^{- x}}

Inverse of

\frac{1000}{100 + 900 e ^{- x}} : - ln (\frac{10 - x}{9 x})

- ln (\frac{10 - x}{9 x})

Find the domain of each inverse function

Domain of

- ln (\frac{10 - x}{9 x}) : 0 < x < 10

Combine the intervals

0 < f (x) < 10

f (x) = \frac{x - 1}{x ^{2} - 25}

f (x) = 4 x^{\frac{3}{5}} - x^{\frac{4}{5}}

(5.3) (4.2)

\frac{x ^{2}}{x ^{2} - 1}

y = - 2 x + 6

What is the range of (1000)/(100+900e^{-x)} ?
The range of (1000)/(100+900e^{-x)} is 0<f(x)<10