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# Introduction to the inverse of a function

Introduction to the inverse of a function. Created by Sal Khan.

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• Why do we need to satisfy two conditions for the inverse function ? i.e. g(f(x)) = Ix AND f(g(x)) = Iy? Is the one not automatically implied if you state the other? Could somebody give me an example of a function that does not satistfy both conditions?
• When g is the inverse of f, for everything in the range of f .. but g also maps some other things too.. and these happen not to be in the domain of f. So then g acts as an inverse to f, but f is not an inverse to g. This can be referred to as a partial inverse, and an actual inverse might be found by restricting the domain of g to the range of f. A concrete example is f is 'principle square root' and g is 'square'.. Doing f first (i.e. take a square root) then doing g (i.e. squaring that number) always gets back to the starting number for anything f can cope with. However, if starting with g, then g(-1) =1 but now f(1) is not -1.
• Can you tell me about difference between liner and inverse functions ?
• A linear function is simply any function that has a straight line as it's graph. That is functions with one x and no degree. A linear function has to do with a single function and tells you what type it is, nothing more.

Inverse functions, on the other hand, are a relationship between two different functions. They can be linear or not. The inverse of a function basically "undoes" the original.

As a simple example, look at f(x) = 2x and g(x) = x/2. To see what I mean, pick a number, (we'll pick 9) and put it in f.
f(9) = 2(9) = 18. Now put this answer in g.
g(18) = 18/2 = 9. You are back to your original number.

Of course, this is a very simple example, they can get much more complicated than this. To find an inverse of a function, you switch the x and y in the original problem and then solve for y to put in function form.
• I don't understand why at Sal says that g dot f maps to the identity matrix of f. Why shouldn't it be the identity matrix of y? If g maps from Y-->X and f from X-->Y shouldn't the composition of these functions be a mapping from Y-->Y?
• gof = g(f) maps from the domain of f (= X) to the codomain of f (= Y), and "g" maps from Y back to X (first you do f, then g - "right to left").
• Hello. I found the inverse of y=x^3, which is y=x^1/3. The problem is, I don't know if it is a function or not. Is it a function? Thanks!
• Indeed, if we look at the cube root function, we see that there is only one value for each x. It is not always the case that a function is invertible, but x cubed is. If we look at x squared, the inverse is (plus or minus) the square root function. This means there are actually two values for every x, therefore the inverse of x squared is not invertible for its entire domain. If we restrict the domain of x squared to values greater or equal to zero and take the inverse (think absolute value), then we will get a function!
• I'm not entirely sure if what he just explained eliminates the possibility that I'll explain below.
Let X and Y be sets.
Let a be an element in X
Let b and c be elements in Y
Let the function g be a mapping from Y to X and when applied to the element b or c in Y it will resolve to the element a in X.
Let the function f be a mapping from X to Y such that the element a in X maps to b in Y.

This creates a situation where element c in Y maps to a in X which maps to b in Y. The identity element of Y applied to c returns c and the identity element of Y applied to b returns b.

My questions are as follows:
Thus it would seem to me that the function f is invertible, and it's inverse is the function g?
Does this mean that the composition of f with g is not the identity element from Y?
What about the composition of g with f? Is that the identity element from X?
Maybe another way of putting my question is: Is g the inverse of f? Is f the inverse of g?

I'm quite confused. Have I just constructed a system that is impossible or at least isn't allowable in linear algebra?
• One of the conditions for a function to be invertible is that is one-to-one, meaning that each element in the domain maps to one (and only one) element in the co-domain. From what you present, `f` appears for fulfil this, but `g` doesn't, since multiple values in it's domain map to the same value in it's co-domain. Therefore `f` might be invertible, but `g` is clearly not invertible, and for that it cannot be the inverse of any function.
• I'm sorry I don't understand. What is an identity?

Also why is a Linear Algebra video in the normal Algebra section?
• The identity function acts like a 1. If you multiply a function by it you get the same answer.
• Can someone help me with this question please?

Let T and L be linear transformations. T: R3 to R2. L: R2 to R3.

a. Prove that the composition LT cannot be invertible.

b. Give an example to show that TL can be invertible.
• What is a transformation?
• Is y=x^2-2 the inverse of y= sqroot (x+2)?

Secondly, how can you show algebraically where these two graphs intersect?