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

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

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

range

f (x) = \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

7 x - y = - 14, (0, 0)

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

f (x) = \frac{2 x ( x + 1 )}{x + 1}

(- 2, - 7), (0, 5)

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

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