# Discrete Math Functions

For f: R -> R defined by

3x - 5, x > 0;
f(x) = {
-3x + 1, x <= 0,

determine f^-1([-5,5]).

Can someone help out and show how to do this. This is a sample problem for a test.
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Commented:
Draw the graph of f(x) first, the inverse becomes obvious. The inverse is multi valued btw.
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Author Commented:
It needs to be in a form of a proof.
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Commented:
Can you give an example of one that you do know how to do so that we can see how it should be structured?

f(x) is one-to-one and linear over x>0 and over x<=0 separately, so the domain that gives the range -5,5 should be easy to find. Then you just set y=f(x) and solve for x (again separately for the different pieces).
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Author Commented:
I'm not sure how to do this at all, but your explanation was very clear. The way the problem is worded to begin with is very confusing as I'm not sure what they are asking you to do. So just to clarify, that problem asks to simply find the domain and solve for x?

If so, do you just solve for x, and then plug in the range [-5, 5] into the y values?

Example 1: y = 3x -5
--> x = (y + 5)/3
--> x = (5 + 5)/3
--> x = 10/3

Example 2: y = -3x + 1
--> x = (y - 1)/-3
--> x = (-5 - 1)/-3
--> x = 2

Still a bit confused as you can tell.
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Commented:
Ah, not quite, but close.

For x>0:   x = (y + 5)/3

For x<=0: x = (y - 1)/-3

But you need to find the ranges separately for each of the two options.
So for x>0
(-5 + 5)/3 = 0
(5 + 5)/3 = 10/3
So the range over [-5,5] is [0,10/3]
But x > 0 so the range of valid answers should be (0, infinity) so the real range is (0, 10/3].

For x<=0
(-5 - 1)/-3 = 2
(5 - 1)/-3 = -4/3
So the range over [-5, 5] is [-4/3, 2]

But x <= 0, so the range of valid answers is (-infinity, 0] so the real range is [-4/3, 0]

I'm not sure how the answer is supposed to look, but I would write it something like this

(y + 5)/3     for a<y<b
f^-1(y) = {       undefined   for c<y<d
(y - 1)/-3     for g<y<h

Some of those '<'s should be '<=' probably.
I'll leave it to you do try to determine the domains. But the typical notion of a function is that it cannot be multivalued so in the cases where you can't tell which f(x) was used, I would call it undefined.

Does that make sense.
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Commented:
Again, all of the answers will be clear in a graph if it's carefully drawn.
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Author Commented:
Almost everything is clear except one piece of terminiology:

"But you need to find the ranges separately for each of the two options."

if you have set A and B where a is a member of A and b is a member of B, and if f(a)=b, then f^-1(b) = a, right?

According to above statement, b is in the range, and a is in the domain.

Then -->according to  f^-1([-5,5])  --> [-5,5] is the range right, and the inverse would be the domain right? Then how come your quote above says that you need to find the RANGES separately. Don't we already know the Range to be [-5,5], and we need to find the DOMAIN not the RANGE?
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Commented:
Well, the domain of f is the same as the range of f^-1

"if you have set A and B where a is a member of A and b is a member of B, and if f(a)=b, then f^-1(b) = a, right?"

In your case, the domain of f is A and the range of f is B but for f^-1 it's flipped. The domain of f^-1 is B and the range is A.
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Author Commented:
Excellent. Can you tell me if the question was worded a bit differently like below, what would change as far as my appraoch goes. No need to solve the below, I'm just curious what happens when you say its no longer R --> R, and now Z --> N. What happens? And now stating it is one-to-one and onto? How does that change my appraoch as well?

The function f: Z --> N defined by
2x - 1, if x > 0;
f(x) = {
-2x, otherwise.

is one-to-one and onto. Fine f^-1.
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Commented:
If it's one-to-one and onto, then the inverse will be defined everywhere on the whole domain.
If f is Z->N, it doesn't really change anything. Just f^-1 will of course be N->Z
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Author Commented:
And since this time they are not providing a domain as in f^-1[-5,5] and instead only asking to find f^-1, all they are asking for is to solve for x in the 2 functions below?

The function f: Z --> N defined by

2x - 1, if x > 0;
f(x) = {
-2x, otherwise.

is one-to-one and onto. Fine f^-1.
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Commented:
Yes. All you need to do is reverse the function. It will have a { and two options just like f did.
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