Does the Function Have an Inverse? A Theorem for Proving Inverse Functions

[iloveannaw]

Homework Statement

Given the function

f(x) = \frac{ax+b}{cx+d}

where f: \mathbb{R} \backslash \left\{ \frac{-d}{c} \right\} \rightarrow \mathbb{R}

show that f is either a constant or has an inverse function.

I can see why this would be true. If a function takes all real numbers and returns all real numbers then it could either be a multiplying constant or have an inverse, e.g. x²
wouldn't work here because it only returns values greater than zero.

My question is is there a theorem or lemma or something that can help me prove this?

thanks

 

[iloveannaw]

ah, i just found out about the bijective requirements for inverse functions – sorry we didn't cover this in class.

but I'm a bit confused injection and surjection seem quite the same. Isn't it enough to show that the function is one-to-one??

 

[HallsofIvy]

No, "injective" and "surjective" are not the same. For example, the cubic function f(x)= (x- a)(x- b)(x- c) is surjective- given any number y, there exist an x such that f(x)= y. That is true because \lim_{x\to\infty}f(x)= \infty, \lim_{x\to-\infty} f(x)= -\infty and f is continuous.

But f is NOT injective: f(a)= f(b)= f(c).

 

[iloveannaw]

Hi, thanks for the reply

Looking more at the given function i can show that it is indeed one-to-one. As to whether it is surjective I'm not really sure what my answer is telling me:

f\left(x\right) = \frac{ax+b}{cx+d} = \xi

ax+b = cx\xi+d\xi

x = \frac{d\xi - b}{a - c\xi}

where \xi, a, b, c, d \in \mathbb{R}. The question states that

cd \neq 0, meaning that either a or b could be zero. If we say:

a = b = 0, then

x = \frac{-d}{c}, which is 'not allowed' as the original set excludes that term. What does this mean?
thanks

 

[HallsofIvy]

If a and b are both 0 then the original function is f(x)= 0, for all x, which has no inverse. Remember, the problem was to show that either f is a constant or has an inverse.