What is the range of (e^x+1)/(e^x-2) ?

The range of (e^x+1)/(e^x-2) is f(x) 1

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range of (e^x+1)/(e^x-2)

range

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

f (x) < - \frac{1}{2} or f (x) > 1

Interval Notation

(- \infty, - \frac{1}{2}) \cup (1, \infty)

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

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

Inverse of

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

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

Find the domain of each inverse function

Domain of

ln (\frac{1 + 2 x}{x - 1}) : x < - \frac{1}{2} or x > 1

Combine the intervals

f (x) < - \frac{1}{2} or f (x) > 1

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

y - 3 x = - 1

f (x) = - 2 x

lo g_{64} (x - x^{2})

f (x) = \frac{1}{( x - 3 ) ^{2}}

What is the range of (e^x+1)/(e^x-2) ?
The range of (e^x+1)/(e^x-2) is f(x)<-1/2 \lor f(x)>1