```Question 616249
{{{f(x)=ln(x+1)+1}}}

{{{y=ln(x+1)+1}}}

{{{x=ln(y+1)+1}}}

{{{x-1=ln(y+1)}}}

{{{e^(x-1)=y+1}}}

{{{e^(x-1)-1=y}}}

{{{y = e^(x-1)-1}}}

So the inverse function is <img src="http://latex.codecogs.com/gif.latex?\LARGE f^{-1}(x) = e^{x-1}-1" title="x" />

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Domain of f(x):

x+1 > 0

x > -1

So the domain is x > -1 which in interval notation is (-1,infinity), which is what you have. Nice job.

Range of f(x):

The range is the set of all real numbers since the domain of the inverse function is the set of all real numbers.
Also graphing f(x) shows us that all y values get hit.

Asymptote for f(x): It's the vertical asymptote x = -1 (since this value is at the boundary of the domain)

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Let g(x) be the inverse of f(x)

Domain of g(x): Set of all real numbers since you can plug in ANY number you want into {{{y = e^(x-1)-1}}} and you'll get some number out.

Range of g(x): The range is y > -1 which in interval notation is (-1,infinity). This is just the domain of f(x). Remember that the two switch.

Asymptote for g(x): It's the horizontal asymptote y = -1 (since this value is at the boundary of the range)

Again, remember that the domains, ranges, and asymptotes all switch when going from a function to its inverse.```