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Here is the question:

(i) State the range of this function:$$(x+2)/(2x+1)$$ Edit: domain $x>0$

(ii) Find the inverse function of $f^-1$

I initially attempted to find the range by calculating the domain of the inverse function:$$x=(y+2)/(2y+1)$$ $$2xy+x=y+2$$ $$y(2x-1)=2-x$$ $$y=(2-x)/2x-1)$$ And found that a domain of $x=1/2$would mean dividing by $0$.

So I found the domain to be $x≠1/2$. Meaning that the range of $f^-1$ is $x≠1/2$.

However I am assuming I cannot do this as part (ii) asks me to find the inverse function, therefore I must be doing another method to find the range?

StrangeApple
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  • You have actually found it: $f^{-1}(x) = {2-x \over 2x-1}$, with domain $\mathbb{R} \setminus { {1 \over 2} }$ and codomain $\mathbb{R} \setminus {-{1 \over 2}}$. – Abstraction Nov 20 '15 at 12:59
  • To found the range, the domain should be given, as the range strongly depends on the domain. So: On what domain do you consider $f(x) = (x+2)/(2x+1)$. – martini Nov 20 '15 at 13:04
  • Since the OP does not state the domain, I guess it will be $\mathbb{R} \setminus {-\frac{1}{2}}$. –  Nov 20 '15 at 13:08
  • I've just realised the the domain for the function is x>0 – StrangeApple Nov 20 '15 at 13:08

1 Answers1

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You calculated the inverse function and then analyzed its domain.

This contained the assumption that whole $\mathbb{R}$ was available for the inverse function, while the original function might have had restrictions on its domain which might have led to restrictions of its range which is the domain of the inverse.

E.g. it could have been stated as $$ f(x) = \frac{x+2}{2x+1} \quad (x < -3) $$

I am not sure if you considered this, however you were lucky as the range of the original function was $\mathbb{R} \setminus \{ 1/2 \}$ which is what your analysis gave as well.

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An alternative method would have been to look at the original function and note that it could be written as $$ \frac{x+2}{2x+1} = \frac{x+ 1/2 + 3/2}{2(x + 1/2)} = \frac{1}{2} + \frac{3}{4(x+1/2)} $$ which is $1/x$ scaled by $3/4$ and the origin shifted to $(-1/2, 1/2)$ and deducing the range from it.

mvw
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