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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.

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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 $$ (x+2)y=2x-1. $$

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  • $\begingroup$ 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
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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.

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