Function: Prove it is 1-to-1, Onto, and find Inverse

[rooski]

Homework Statement

Let a,b,c E R with b != ac and let the function f : R --> R be given by

f(x) = a if x = c
f(x) = (ax - b) / (x - c) if x != c


Show that f(x) is one-to-one
Show that f(x) is onto
Show the inverse of f(x)

The Attempt at a Solution

I don't want anyone to solve this for me, i just need a push in the right direction. My textbook is next to useless. There are no similar examples showing how to work with piecewise functions.

The one thing my book shows is that we take 2 arbitrary variables and prove that f(v1) and f(v2) have differing values in the domain and range.

How can you even conclude that a function is one-to-one with such a weak proof though? Especially complex piecewise functions like mine.

 

[Mark44]

Homework Statement

Let a,b,c E R with b != ac and let the function f : R --> R be given by

f(x) = a if x = c
f(x) = (ax - b) / (x - c) if x != c


Show that f(x) is one-to-one
Show that f(x) is onto
Show the inverse of f(x)

The Attempt at a Solution

I don't want anyone to solve this for me, i just need a push in the right direction. My textbook is next to useless. There are no similar examples showing how to work with piecewise functions.

The one thing my book shows is that we take 2 arbitrary variables and prove that f(v1) and f(v2) have differing values in the domain and range.

Your book should have a better definition of one-to-one than this - something like f(x1) = f(x2) ==> x1 = x2. (I switched from your v1 and v2 to x1 and x2.)

This is equivalent to x1 $\neq$ x2 ==> f(x1) $\neq$ f(x2).

Start with two different x values, x1 and x2, and show that f(x1) $\neq$ f(x2). Since you have a piecewise defined function, I think you'll need two different cases:
1. x1 = c, x2 $\neq$ c
2. x1 $\neq$ c, and x2 $\neq$ c, with x1 and x2 being different.

I don't think you need a third case, with x1 $\neq$ c, and x2 = c, since that would be covered by case 1.

That's how I would start out.

How can you even conclude that a function is one-to-one with such a weak proof though? Especially complex piecewise functions like mine.

 

[rooski]

Okay with those cases in mind I'll start at Case 1.

f(x1) = f(x2) translates to a = (ax-b)/(x-c)

Right?

I am confused because in my notes, they essentially whittle the equation down until it shows x1 = x2, but only one side of my equation has x in it.

 

[Mark44]

Okay with those cases in mind I'll start at Case 1.

f(x1) = f(x2) translates to a = (ax-b)/(x-c)

Right?

You're going at it the opposite way that I suggested, which was to show that, if x1 != x2, then f(x1) != f(x2).

I am confused because in my notes, they essentially whittle the equation down until it shows x1 = x2, but only one side of my equation has x in it.

 

[rooski]

Oh sorry. So i want to aim to prove that a != (ax-b)/(x-c)...

Isn't that already painfully obvious, just from looking at it though? Is there any simplifying needed for this?

 

[Mark44]

Just looking at it wouldn't be enough. Since it's so obvious, it should be easy to show that if x1 != x2 (and as in case 1), then f(x1) != f(x2). After evaluating f(x2) the resulting expression should involve x2.

 

[rooski]

I kind of shuffled case 1 to the side - i don't exactly know how to go about doing it. I got some progress on Case 2.

-snip, error-

 

[Mark44]

So what i did now was firstly remove c from the bottom

What's your justification for doing that?

 

[rooski]

-EDIT-

i figured out Case 2.

I'm stuck on case 1 though.

a != (ax-b)/(x-c)

Since this is an inequality, what manipulations can i do? Can i move a to the right hand side? I want to eventually prove that both sides can never have the same value, right?

 

[rooski]

Alright i finished the question, thanks for the help all.

 

[Confused31]

Alright i finished the question, thanks for the help all.

What was the answer? I am trying to work on the exact same question and I am totally confused.

 

[rooski]

To find if it is 1-to-1 you must break it into 2 cases, which i discussed earlier.

The trick is to use the fact that b-ac != 0 in order to show x1 = x2.

To show the function is onto, you examine both cases again. Case 1 is simple, and Case 2 you must solve in terms of x (invert it so that you get x = equation with y in it).

 

[Confused31]

To find if it is 1-to-1 you must break it into 2 cases, which i discussed earlier.

The trick is to use the fact that b-ac != 0 in order to show x1 = x2.

To show the function is onto, you examine both cases again. Case 1 is simple, and Case 2 you must solve in terms of x (invert it so that you get x = equation with y in it).

im still lost, can you give me a little more info to help push me in the right direction.