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For $a,b,c,d \in \mathbb{R}$ and $ad-bc \neq 0$ I have the function $f : \mathbb{R}\setminus{\{d/c\}} \to \mathbb{R}, f(x) = \frac{ax-b}{cx-d}$. How can I find the largest possible restriction of $f$ such that $f$ is injective on it and the image of this restriction? Thank you for your help.

Edit: If I understand correctly, the linked question is concerned with finding a condition for injectivity on the parameters a,b,c,d (the condition they find is $bc-ad \neq 0$). With mine the values are already given.

Also I know that (as a continuous function) it's injective iff it is strictly monotone, which is the case when $f'(x) = \frac{bc-ad}{(cx-d)^2}$ is positive or negative. But I don't understand how this helps me, since the denominator is always positive and the numerator a fixed value for given bc and ad? If this is even the right approach for finding a restriction.

firstsnow
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1 Answers1

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Your computation of the derivative shows that if $ad-bc \neq 0$ then $f$ is strictly monotone on each side of $d/c$. So $f$ is injective on each side of $d/c$; you can then separately check that it is injective on its full domain.

To examine surjectivity, notice that by the piecewise monotonicity, the image will be the union of the interval with endpoints $\lim_{x \to -\infty} f(x)$ and $\lim_{x \to d/c^-} f(x)$ and the interval with endpoints $\lim_{x \to \infty} f(x)$ and $\lim_{x \to d/c^+} f(x)$. Compute these four limits.

Another way to look at it is to write $f(x)=a/c+k/(cx-d)$, which basically means that injectivity/surjectivity can be studied just for $g(x)=1/x$.

Ian
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