I just encountered a new type of question in my textbook that I'm not sure how to do. It says:

The one-one function $f$ is defined on the domain $x>0$ by $f(x)=\frac{2x-1}{x+2}$

I've been asked to find the range, $A$, of $f$ and obtain an expression for the inverse $f^-1(x)$, for $x∈A$

Would someone please explain to me how to do this?

Thank you!

Edit: I know how to find the inverse, it's the range $A$ that's troubling me.


Since the given function is continuous and bijective, you know that its range will be from $\inf_{(0,+\infty)} f$ to $\sup_{(0,+\infty)} f$, possibly including or excluding these two particular values. In this case, the derivative of $f$ is always positive, so it attains its inferior bound for $x\to 0^+$, that is $-\frac12$, while its superior bound is given by $x\to+\infty$ and equals $2$; $f$ never actually reaches it, but it has a horizontal asymptote at this value.

Given this information, the range is $\bigl(-\frac12,2\bigr)$.

You can find the inverse simply by putting $y=f(x)$ and solving for $x$, i.e. solving the equation $$ (x+2)y=2x-1. $$

  • $\begingroup$ Observe that $\;f(-7)=\frac{-15}{-5}=3\notin\left(-\frac12,\,2\right)\;$ $\endgroup$ – DonAntonio Mar 12 '16 at 20:55
  • $\begingroup$ The function is defined on the domain $(0,+\infty)$. $\endgroup$ – yellowquark Mar 12 '16 at 20:56
  • $\begingroup$ I don't quite understand the inf and sup bits and the derivative that you mentioned, can you please explain it to me? $\endgroup$ – Spica Mar 12 '16 at 21:02
  • $\begingroup$ By Darboux's theorem, the range of $f$ is an interval $(a,b)$, and (I would say this is by definition) $a=\inf_{(0,+\infty)} f$ and $b=\sup_{(0,+\infty)} f$. I used the derivative to see if $f$ had some extrema: since $f'(x)\ne 0$ for all $x\in(0,+\infty)$, those extrema are found at the boundary. Lastly, $f'>0$, so the lower bound is at $x\to 0^+$ while the upper bound at $x\to+\infty$, because $f$ is increasing. $\endgroup$ – yellowquark Mar 12 '16 at 21:08

Note that the inverse map $f^{-1}$ satisfies $f^{-1}(f(x)) = x$. For some $x >0$ say $y = f(x)$. Then $$y= f(x) = \frac{2x - 1}{x+2} = \frac{2f^{-1}(f(x)) - 1}{f^{-1}(f(x)) + 2} = \frac{2f^{-1}(y) - 1}{f^{-1}(y) + 2}.$$ Now you can solve for $f^{-1}(y)$ which will tell you what the inverse map looks like.

As for finding the range, you need to find all values $y \in \mathbb R$ such that there is an $x > 0$ with $f(x) = y$. Another post has already addressed this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.