Inverse of a Piecewise Function

[SpringPhysics]

Homework Statement

Find the inverse of
...{x...x =/= a₁,...,a_n
f(x) = {a_i+1...x = a_i, i = 1,...,n-1
...{a₁...x = a_n

Homework Equations

The Attempt at a Solution

I interchanged the variables x and y, but I am very confused as to how to solve for the rest. I don't understand how to find inverses if we aren't given an explicit formula. Can someone help please?

 

[LCKurtz]

Homework Statement

Find the inverse of
...{x...x =/= a₁,...,a_n
f(x) = {a_i+1...x = a_i, i = 1,...,n-1
...{a₁...x = a_n

Homework Equations

The Attempt at a Solution

I interchanged the variables x and y, but I am very confused as to how to solve for the rest. I don't understand how to find inverses if we aren't given an explicit formula. Can someone help please?

It might help you to just list the ordered pairs for f. Then reverse them all and you should see a way to write f^-1 as a formula similar in form to the formula for f(x).

 

[SpringPhysics]

It might help you to just list the ordered pairs for f. Then reverse them all and you should see a way to write f^-1 as a formula similar in form to the formula for f(x).

All right. Thanks for your help.

Also, could I ask for help on finding inverses for the following?

f(x) = x + [x] (floor) and f(x) = x/(1-x²) for -1 < x < 1.

For the first function, there is no reversible equation for the floor operator, so could I state the inverse as simply {(x+[x],x) | (x,x+[x]) \in f}?
Would it be possible to state that any x in f be a.b, where a is an integer and b is any real number?
Then
f(x) = a.b + a for a >= 0 and
f(x) = a.b + (a-1) for a < 0.
Then the inverse of f would be given by
f^-1(x) = 1/2 (x + 0.b) for x >= 0 and
f^-1(x) = 1/2 (x + 1.b - 0.(2b)) for x < 0.

For the second function, I interchanged the variables and obtains:
x(1-y²) = y
0 = xy² + y - x
Using the quadratic formula, I got
y = (-1 +/- \sqrt{1+4x^2})/2x, -1 < y < 1
How do I know whether to take the positive or negative?

 

[HallsofIvy]

You shouldn't take either one. If your calculations are correct, you are saying that this function is NOT one to one and so does NOT have an inverse.

 

[SpringPhysics]

You shouldn't take either one. If your calculations are correct, you are saying that this function is NOT one to one and so does NOT have an inverse.

I am not sure which function you are referring to, but for the second function, the question asked to determine f^-1 (the inverse) for -1 < x < 1. Hence, the function is 1-1 for the specified interval. I am just not sure there is an intuitive reason why the positive root works but not the negative.

The floor function is not 1-1 so it does not have an inverse, but the question is f(x) = x + floor (x), so that the function is 1-1. However, I am not sure how to cleanly express its inverse.

EDIT: Never mind for the floor function. Can someone please explain the first function?

 

[LCKurtz]

EDIT: Never mind for the floor function. Can someone please explain the first function?

Are you talking about f(x) = x + [x]?

Try drawing the graph of f^-1. You will see that its domain has gaps and the segments of the graph are translates of f(x) = x. Does that help?

 

[SpringPhysics]

Are you talking about f(x) = x + [x]?

Try drawing the graph of f^-1. You will see that its domain has gaps and the segments of the graph are translates of f(x) = x. Does that help?

Sorry, I meant f(x) = x/(1-x²)

 

[LCKurtz]

When you interchanged x and y and solved for y you got:

y = \frac {-1 \pm \sqrt{1+4x^2}}{2x}

One way you can tell you don't want the - choice is what happens as x approaches 0. The branch you want goes through the origin. If you look at

y = \frac {-1 - \sqrt{1+4x^2}}{2x}

as x\rightarrow 0 you get a -2/0 form which indicates a vertical asymptote.

On the other hand if you let x\rightarrow 0 in

y = \frac {-1 + \sqrt{1+4x^2}}{2x}

you get 0 as you can see if you rationalize the numerator and take the limit.

Another thing that is a bit more work is to observe that with the + choice you get -1\le y \le 1, which also tells you you have the right branch.

 

[SpringPhysics]

I see now. Thank you so much for your help!