Proving that a function is not Surjective.

UserKunal123

New member
This is the question from my text book:
Show that f:R-{3}—>R defined by f(x)=(x-2)/(x-3) is not onto. This is what I have done so far:

f(x)=y=(x-2)/(x-3)
=> x=(3y-2)/(y-1)
For y=1, x is undefined.

This means for y=1 there is no pre-image x in the Domain. Hence, the function is not onto.

I doubt that the function is not onto just because for y=1, x is undefined. What have I done wrong and how do I prove that the function is not onto?

Dr.Peterson

Elite Member
I think you've done well. Nothing is wrong.

If you look at the graph of this function, you will see that while its domain is all reals except 3, its range is all reals except 1, just as you've found. There is no value of x for which y = (x-2)/(x-3) is 1.

Jomo

Elite Member
What you did is fine.
Here is how I solve these problems.
Let k be in R.
Now solve (x-2)/(x-3) - k for x. So x = x=(3k-2)/(k-1). Now f((3k-2)/(k-1)) = k except for when x =(3k-2)/(k-1) is undefined. This happens when k=1. So 1 is not in the image of f and f is therefore not onto.

I admit that this 99.9% of what you did. I just like picking a element in the range (k) and then show that there is a restriction on k ( k is not 1), so f is not onto or to show that there is no restriction on k, so f is onto..