1
$\begingroup$

The question:

Show that the function $f : \mathbb{R} − \{−1\} → \mathbb{R} − \{2\}$ defined by $f(x) = \dfrac{4x + 3}{2x + 2}$ is a bijection, and find the inverse function.

How would I establish the function is a bijection, and how do I find the inverse using discrete math?

I understand a bijection means the function must both be onto and one-to-one, which it is, but how do you prove this?

As for the inverse, using algebra I have determined it to be $\dfrac{3-2x}{2x-4}$, but am I supposed to solve it some other way?

$\endgroup$
  • $\begingroup$ Please, firstly establish your latex. $\endgroup$ – Jihad Oct 26 '14 at 3:46
  • $\begingroup$ Edited, I think that's everything $\endgroup$ – user3362196 Oct 26 '14 at 3:53
2
$\begingroup$

The inverse function $g(x)$ should satisfy that $f\circ g = g\circ f =$ identity map. If such $g$ exists, then automatically $f$ is a bijection.

Now that you have already calculated the inverse, then check that the above condition is valid, and you are done.

Identity map is a function that sends $x$ to $x$ for every $x$ in its domain. In your case, $f\circ g$ means $x\mapsto g(x)\mapsto f(g(x))$ i.e. $x\mapsto\cfrac{3-2x}{2x-4}\mapsto f\left(\cfrac{3-2x}{2x-4}\right)=x$. This shows that $f\circ g$ is the identity map. Maybe try to prove $g\circ f$ by yourself?

$\endgroup$
  • $\begingroup$ We were never taught about that map, can you link me to a site that explains it? Or if you are willing to explain it that would be great $\endgroup$ – user3362196 Oct 26 '14 at 4:03

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.