MHB What is the Domain for the Inverse of a One-to-One Function?

[ardentmed]

Hey guys,

I've a few more questions this time around from my problem set:

(Ignore question 2abc, I only need help with the first one)

Question:

For the first one, in order to prove that a function is one-to-one, f(x1) =/ f(x2) when x1 =/ x2. Thus, the horizontal test applies. So I said that:

(e^a)/(1+4^a) = (e^b)/(1+4^b)
This gave me (e^a)=(e^b)
Hence, a = b and f is one to one.

As for the inverse, I wrote out y=f(x), solved for x in terms of y, and expressed f^-1 as a function of x where x and y interchange. Ultimately, y=(e^x)/(1+4e^x) ended up becoming f^-1(x) = ln(-x/(4x-1))

As for the domain, I know that x< 0 to avoid non-real answers to comply with ln.

Hence the domain is (-infinity, 0)u(1/4, infinity). Which I highly doubt. Any ideas?
Thanks again. I really appreciate the help guys.

 

[Euge]

Re: Inverse One to One Proofs Help Please

Hey guys,

I've a few more questions this time around from my problem set:

(Ignore question 2abc, I only need help with the first one)

Question:For the first one, in order to prove that a function is one-to-one, f(x1) =/ f(x2) when x1 =/ x2. Thus, the horizontal test applies. So I said that:

(e^a)/(1+4^a) = (e^b)/(1+4^b)
This gave me (e^a)=(e^b)
Hence, a = b and f is one to one.

This part you have right.

As for the inverse, I wrote out y=f(x), solved for x in terms of y, and expressed f^-1 as a function of x where x and y interchange. Ultimately, y=(e^x)/(1+4e^x) ended up becoming f^-1(x) = ln(-x/(4x-1))

As for the domain, I know that x< 0 to avoid non-real answers to comply with ln.

Hence the domain is (-infinity, 0)u(1/4, infinity). Which I highly doubt. Any ideas?
Thanks again. I really appreciate the help guys.

Your answer is close. While the inverse function is correct, the domain is off; it should be $(0, 1/4)$. The domain of $f^{-1}$ is the set of points $x$ such that $x/(1 - 4x) > 0$. So $x$ belongs to the domain of $f^{-1}$ if and only if

(1) $x > 0$ and $1-4x > 0$, or

(2) $x < 0$ and $1-4x < 0$.

In case (1), $x > 0$ and $x < 1/4$, which is equivalent to saying $x \in (0,1/4)$. In case (2), $x < 0$ and $x > 1/4$, which cannot both happen. So the domain of $f^{-1}$ is the set of points in (1), which is (0, 1/4).