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## Algebra 1

### Course: Algebra 1>Unit 8

Lesson 13: Intro to inverse functions

# Inputs & outputs of inverse functions

Sal explains that if f(a)=b, then f ⁻¹(b)=a, or in other words, the inverse function of f outputs a when its input is b.

## Want to join the conversation?

• I have a doubt. An x in the domain of a function is mapped to just one y in the range. But y could be mapped from more than one x. So, what is the result of the inverse of a function when you input a y that could be mapped to more x? Is it possible? Will the function return more results? Or do we just swap x and y and do we still have more x mapping to a single y? Thanks for your time! • @ the result of f^-1(f^-1(13)) was found to be 9. From f^-1(13)=5 and then f^-1(5) = 9. I am confused because I thought that it would have been 13 again, just as the inverse of 7 example. I am having a hard time reconciling this issue. • @ I don't understand how you got the inverse of 7. • What is many-to-one and one-to-one? • At about to , Sal mentions that the function is a one-to-one mapping, since no two x's map to the same f(x). This means the function is invertible (has an inverse).
So if we're given a value for f(x), we will know without doubt which value of x produced it.
A many-to-one mapping means that at least two values of x (and maybe more) map to a single value of f(x).
So if we were given a value of f(x) to start with, we wouldn't be able to say with certainty which value of x had produced it.
• Is it possible to find F^-1(f(58)) with a table of
x 5, 3, 1, 18, 0, 9
f(x) 9, -2, -5, -1, 1, 11?
I don't think it is but I was asked this in a problem and was wondering if this could a mistake.
When I looked at the answer it said it was 58. Why is this? • This is true by definition of inverse. f(58) would lend an answer of (58,y) depending on the function. It really does not matter what y is. The inverse of this function would have the x and y places change, so f-1(f(58)) would have this point at (y,58), so it would map right back to 58.
So try it with a simple equation and its inverse. If f(x)=2x + 3, inverse would be found by x=2y+3, subtract 3 to get x-3 = 2y, divide by 2 to get y = (x-3)/2. Lets find f-1(f(4)). f94) = 2(4)+3 = 11. So f-1(11) = (11-3)/2 = 8/2= 4.
or f-1(f(-5) f(-5) = 2(-5) + 3 = -7, f-1(-7) = (-7-3)/2 = -10/2 = -5. Try f-1(f(58)). f(58) = 2(58)+3=119. f-1(119) = (119-3)/2 = 116/2 = 58. So the table is irrelevant to the question, it would work for any function.
• This explanation is so much clearer than the intro to inverse functions. The key here is the tables, and that the roles of x and y are reversed. That explains why the inverse graph does not overlay the original function graph in the intro. • what if we had the table below:
x |f(x)
1 | 3
3 | 5
7 | 9
8 | 5

as you see 5 is repeating for different x values {3,8}.
in this case is f^-1(5) undefined ? • what if one of the values isn't on the table, for example, i have the same exact problem as Sal, but it is Inverse f ( f(576)) ? so how would you do this?   