SOLUTION: Find the inverse of the following function. Find domain, range, and asymptotes of each function. f(x)=ln(x+1)+1 I tried to solve this but i got mixed up. Here is what I had.

Question 616249: Find the inverse of the following function. Find domain, range, and asymptotes of each function. f(x)=ln(x+1)+1
I tried to solve this but i got mixed up.
Here is what I had.
y=(x+1)+1
x=(y+1)+1
x-1=(y+1)-1
x-2=y
x-2=f^-1(x)
My teacher noted that I didn't add the ln to the equation upon solving.
Would it be x=ln(y+1)+1
Can someone please help me. Thanks
Also for the Domain, I'm guessing that I solve
x+1=0
x=-1 so the domain is (-1,inf)
I thought that there cannot be a domain for x<0.
So confused
Answer by jim_thompson5910(28717) (Show Source):
You can put this solution on YOUR website!

So the inverse function is $f^{-1}(x) = e^{x-1}-1$
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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 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.