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Let $f(x) = \dfrac{x + 2 }{x - 3}$.

There's three parts to this question:

  1. Find the domain and range of the function $f$.
  2. Show $f^{-1}$ exists and find its domain and range.
  3. Find $f^{-1}(x)$.

I'm at a loss for #2, showing that the inverse function exists. I can find the inverse by solving the equation for $x$, showing that it exists without just solving for the inverse. Can someone point me in the right direction?

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I am also at a loss, because you told us how to solve 2: " I can find the inverse by solving the equation for x". That's it... To find the inverse, just solve $f(x)=y$ for $x$. If you can find the inverse it exists. –  N. S. Oct 25 '11 at 19:40
    
For #2, perhaps you were expected to show that $f$ is invertible by showing that it is bijective (injective and surjective). –  Srivatsan Oct 25 '11 at 19:46
    
Shriva: In calculus, "invertible" usualy means 1-1. This can be proven by showing that $f' < 0$, but that would be overkill... eri: if you want to prove that $f$ is 1-1, try using $f(x)=f(x) = \dfrac{x -3+5 }{x - 3}$ –  N. S. Oct 25 '11 at 21:42

3 Answers 3

up vote 3 down vote accepted

Can you graph it? Passing the horizontal line test would show it has an inverse. That may be what your teacher is looking for.

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I believe this is what I'm looking for, I'll select this as the answer in a few minutes when it lets me. Thanks. –  erimar77 Oct 25 '11 at 19:46
    
Except, you have to find the inverse as well so this doesn't help. Finding the inverse shows that it exists. So, there's no need to show it exists another time. –  Graphth Oct 25 '11 at 20:44
    
Except that the question specifically asks for this to be done separately, in a way that reveals the domain and range. And I can see why. At this level I would expect many students to be able to "find the inverse" of $f(x)=x^2$ if they did not specifically investigate whether it exists first. –  AMPerrine Oct 25 '11 at 21:47

A function will be invertible (on a proper domain) if and only if it is injective, that is, if it never takes the same value twice. If your function is continuous and is defined on an interval, this forces it to be increasing or decreasing. Even without being defined on an interval we can use calculus to determine if the function is increasing/decreasing, which lets us solve the problem fairly simply.

We have that $f'(x)=\frac{(x-3)-(x+2)}{(x-3)^2}=\frac{-5}{(x-3)^2}$. This is negative except where it is undefined, and so the function is decreasing, at least locally. Unfortunately, because we have the discontinuity at $x=3$, we can't say immediately that the function is injective. However, we do know that it never takes the same value twice on either side of the discontinuity. This combined with limits, is all we need:

Since $\displaystyle \lim_{x\to -\infty}f(x)=\lim_{x\to \infty}f(x)=1$, we must have that $f(x)<1$ if $x<3$ and $f(x)>1$ if $x>3$. Since no number is both less than 1 and greater than 1, we can't have $f(x_1)=f(x_2)$ for $x_1<3$ and $x_2>3$. Therefore $f(x)$ is injective, and hence invertible on a suitable domain (namely on $(-\infty,3)\cup (3,\infty)$.)

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It is a valid way to find the inverse by solving for x first and then verify that $f^{-1}(f(x))=x$ for all $x$ in your domain. It is quite preferable to do it here because you need it for 3. anyways.

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I would assume there is an obvious answer, as it wouldn't be included as an additional question immediately after the fact. –  erimar77 Oct 25 '11 at 19:46

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