Confusion over inverse of a function (involution)

[Dr.Peterson]

I thought about that but then made the mistake of multiplying the whole thing by 1/3 when it should be (x+3)/2 * 1/3 ( I'm stepping on all the landmines today!) so doing that we're left with (x+3)/6 but comparing that to the original functions and inverses I'm not sure what to conclude.

You've got it.

Now let's check. We're claiming that this is g^-1(f^-1(x)), where f(x) = 2x+1 and g(x) = 3x-2, so that f^-1(x) = (x-1)/2and g^-1(x) = (x+2)/3.

Suppose that x = 9. Then f^-1(9) = 4, and g^-1(4) = 2. So g^-1(f^-1(x)) = 2.

What does your expression give? (x+3)/6 = (9+3)/6 = 2. So it agrees, at least for this input.

Also, the claim is that this is the inverse of the function f(g(x)). So if we put 2 into what you call fg, we should get that 9 back. Do we? Well, g(2) = 4, and f(4) = 9, so that works, too.

There are a lot of other things you might do, if you want to look at the problem from all angles. You might find what f(g(x)) is and work out its inverse, to see that you get the same result, for example.

This sort of "above and beyond" work is the way to milk an exercise of all its value, using it to gain lots of experience in doing the algebra, and also getting opportunities to observe how inverses and complements work out.

 

[Simonsky]

You've got it.

Now let's check. We're claiming that this is g^-1(f^-1(x)), where f(x) = 2x+1 and g(x) = 3x-2, so that f^-1(x) = (x-1)/2and g^-1(x) = (x+2)/3.

Suppose that x = 9. Then f^-1(9) = 4, and g^-1(4) = 2. So g^-1(f^-1(x)) = 2.

What does your expression give? (x+3)/6 = (9+3)/6 = 2. So it agrees, at least for this input.

Also, the claim is that this is the inverse of the function f(g(x)). So if we put 2 into what you call fg, we should get that 9 back. Do we? Well, g(2) = 4, and f(4) = 9, so that works, too.

There are a lot of other things you might do, if you want to look at the problem from all angles. You might find what f(g(x)) is and work out its inverse, to see that you get the same result, for example.

This sort of "above and beyond" work is the way to milk an exercise of all its value, using it to gain lots of experience in doing the algebra, and also getting opportunities to observe how inverses and complements work out.

Many thanks for helping me through that process, I think it has helped me appreciate the concepts. You've also reminded me it's better to 'go the extra mile' with these questions and not just try to do them as quickly as possible without allowing oneself to absorb the process. I need to stop rushing at things, panicking and not allowing myself time.

Your time and insights are much appreciated!

 

[Simonsky]

Another functions question that seems to have stumped me somewhat:

Use the information given to find g. In each case g is a linear function, g(x) = mx+c, for some numbers m and c.

f(x) = 6x (fg)-1(x) = (x+30)/18

So, I first thought: this is the inverse of g (the bit you have to find) plugged into the inverse of f which we know is 6x.

I then worked out the inverse of f(x) as x/6. So g plugged into this should produce (x-30)/18.

Before I go on -am I right so far or has the plot already been lost?

 

[Dr.Peterson]

Another functions question that seems to have stumped me somewhat:

Use the information given to find g. In each case g is a linear function, g(x) = mx+c, for some numbers m and c.

f(x) = 6x (fg)-1(x) = (x+30)/18

So, I first thought: this is the inverse of g (the bit you have to find) plugged into the inverse of f which we know is 6x.

I then worked out the inverse of f(x) as x/6. So g plugged into this should produce (x-30)/18.

Before I go on -am I right so far or has the plot already been lost?

Did you mean this??

f(x) = 6x
(fg)^-1(x) = (x+30)/18
​

The previous problem pointed out (though you never called attention to it, so I'm not sure whether you were taught it separately) that (fg)^-1 = g^-1f^-1, not f^-1g^-1. So you wouldn't plug g^-1 into f^-1.

There are several things you could do here, depending on what you have learned. The most straightforward might be to find fg, the inverse of (x+30)/18, and then apply f^-1 to that.

 

[Simonsky]

Did you mean this??

f(x) = 6x
(fg)^-1(x) = (x+30)/18
​

The previous problem pointed out (though you never called attention to it, so I'm not sure whether you were taught it separately) that (fg)^-1 = g^-1f^-1, not f^-1g^-1. So you wouldn't plug g^-1 into f^-1.

There are several things you could do here, depending on what you have learned. The most straightforward might be to find fg, the inverse of (x+30)/18, and then apply f^-1 to that.

Yes-that's what I meant (couldn't do the -1 superscript). So I've got confused by the order again, maybe I was thinking of f-1(g-1)(x) which was why I wanted to plug the g-1 into the f-1.

I think I'm confused by: (fg)^-1 = g^-1f^-1, not f^-1g^{-1 and why it is the latter rather than the former-so I need to clear that up in my head first! (can't escape the superscript typeface now!).}

 

[Dr.Peterson]

Yes-that's what I meant (couldn't do the -1 superscript). So I've got confused by the order again, maybe I was thinking of f-1(g-1)(x) which was why I wanted to plug the g-1 into the f-1.

I think I'm confused by: (fg)^-1 = g^-1f^-1, not f^-1g^-1 and why it is the latter rather than the former-so I need to clear that up in my head first! (can't escape the superscript typeface now!).

To escape from superscript, just hit the button again -- it toggles on and off. I did that for you above.

This is why I hinted that I'd like to know whether you've been formally taught it, or whether they just pointed it out in passing here. I would hope they would have at least stated, if not explained, these ideas, before making you do things that tempt you to assume the wrong thing.

Have you done any work with the algebra of composition of functions? Composition is not commutative, so you can't change the order; but you can move parentheses around:

(fg)(g^-1f^-1) = f(g g^-1)f^-1 = f i f^-1 = ff^-1 = i

(g^-1f^-1)(fg) = g^-1(f^-1f)g = g^-1 i g = g^-1g = i
​

where I'm using i to represent the identity function, i(x) = x. I used the fact that any function composed with its inverse, gg^-1, is the identity. This shows that g^-1f^-1 is the inverse of fg.

If working with functions alone is confusing, you can do the same with the x:

(fg)(g^-1f^-1)(x) = f(g(g^-1(f^-1(x))) = f(f^-1(x)) = x
​

because g(g^-1(anything) = anything.

Essentially, this is the same idea as in solving an equation: to undo a series of operations, you undo each of them, in reverse order, like removing your shoes before your socks, having put the socks on first. You undo the addition before the multiplication in solving 2x+5 = 7, because in evaluating you do the multiplication before the addition.

 

[Simonsky]

To escape from superscript, just hit the button again -- it toggles on and off. I did that for you above.

This is why I hinted that I'd like to know whether you've been formally taught it, or whether they just pointed it out in passing here. I would hope they would have at least stated, if not explained, these ideas, before making you do things that tempt you to assume the wrong thing.

Have you done any work with the algebra of composition of functions? Composition is not commutative, so you can't change the order; but you can move parentheses around:

(fg)(g^-1f^-1) = f(g g^-1)f^-1 = f i f^-1 = ff^-1 = i

(g^-1f^-1)(fg) = g^-1(f^-1f)g = g^-1 i g = g^-1g = i
​

where I'm using i to represent the identity function, i(x) = x. I used the fact that any function composed with its inverse, gg^-1, is the identity. This shows that g^-1f^-1 is the inverse of fg.

If working with functions alone is confusing, you can do the same with the x:

(fg)(g^-1f^-1)(x) = f(g(g^-1(f^-1(x))) = f(f^-1(x)) = x
​

because g(g^-1(anything) = anything.

Essentially, this is the same idea as in solving an equation: to undo a series of operations, you undo each of them, in reverse order, like removing your shoes before your socks, having put the socks on first. You undo the addition before the multiplication in solving 2x+5 = 7, because in evaluating you do the multiplication before the addition.

Sorry for delayed response to your much appreciated help - to answer your question: I'm self-teaching at present using a textbook designed for 16 year olds (British Curriculum known as G.C.S.E )-I'm actually a 58 year old who has not done any maths for 40 years (!) hence my stumbling and staggering. Of course the maths book I follow is designed for classroom use so it's thanks to forums like this and people like you who are generous enough to help that I can get over the 'stuck' moments when I grind to a halt.

To get back to the question: This is the first time I've done 'composition of functions' ( at least since the age of 18!). Trying to get into maths now is a 'tall order'( Brit idiom?) but I'm persistent ( if nothing else).)

So, thanks for reminding me that the previous question dealt with the issue of (fg)^-1(x) = g^-1f^-1(x)-I got confused because the first one involves inverse of g plugged into the f and the second f -1plugged into g-1. I'm not clear why that works unless it an involution which it wasn't in the case of that question where: f(x) = 2x+1 and g(x) = 3x-2.

For example: fg(x) = 2(3x-2) +1 = 6x -3 so (fg)-1 will be: (x+3)/6

and: g-1f-1(x) will be ((x-1/2) +2)/3 =(x+3)6 but as I said, these are not involution so I don't quite get it- hitting cognitive wall again on that!

 

[Dr.Peterson]

This is the first time I've done 'composition of functions' ( at least since the age of 18!). Trying to get into maths now is a 'tall order'( Brit idiom?) but I'm persistent ( if nothing else).)

So, thanks for reminding me that the previous question dealt with the issue of (fg)^-1(x) = g^-1f^-1(x)-I got confused because the first one involves inverse of g plugged into the f and the second f -1plugged into g-1. I'm not clear why that works unless it an involution which it wasn't in the case of that question where: f(x) = 2x+1 and g(x) = 3x-2.

For example: fg(x) = 2(3x-2) +1 = 6x -3 so (fg)-1 will be: (x+3)/6

and: g-1f-1(x) will be ((x-1/2) +2)/3 =(x+3)6 but as I said, these are not involution so I don't quite get it- hitting cognitive wall again on that!

The fact that (fg)^-1(x) = g^-1f^-1(x) is not true only of involutions, but of inverses and composition in general. You've demonstrated it here.

If I were working with you in person, I would have grabbed your book and looked through it, starting in the index, to see whether it actually teaches this fact, or just mentions it in passing and expects you to figure it out (or at least not take incorrect shortcuts). I wouldn't expect it to assume you already know it, whether from last year or 20 years ago. I'd also want to see whether anything they do depends on this; it may just be a distraction. Or it may be something I'd point out for you to read.

But let's look at (fg)^-1 closely, to see what is happening. Take fg in the unsimplified form y = 2(3x-2) +1 and invert that by solving for x, without ever simplifying:

y = 2(3x - 2) + 1
y - 1 = 2(3x - 2)
(y - 1)/2 = 3x - 2
(y - 1)/2 + 2 = 3x
((y - 1)/2 + 2)/3 = x

That's exactly the unsimplified form of g^-1f^-1(x)! Notice that in solving, I undid one operation at a time, starting on the "outside": I subtracted 1, divide by 2, added 2, and divided by 3. That is, I first performed the inverse of f, and then performed the inverse of g. Because f is "on the outside", I undid that first, so that f^-1 is now "on the inside".

Another meaning of the term "invert" is to "turn inside out". That's what's happening!

 

[Simonsky]

The fact that (fg)^-1(x) = g^-1f^-1(x) is not true only of involutions, but of inverses and composition in general. You've demonstrated it here.

If I were working with you in person, I would have grabbed your book and looked through it, starting in the index, to see whether it actually teaches this fact, or just mentions it in passing and expects you to figure it out (or at least not take incorrect shortcuts). I wouldn't expect it to assume you already know it, whether from last year or 20 years ago. I'd also want to see whether anything they do depends on this; it may just be a distraction. Or it may be something I'd point out for you to read.

But let's look at (fg)^-1 closely, to see what is happening. Take fg in the unsimplified form y = 2(3x-2) +1 and invert that by solving for x, without ever simplifying:

y = 2(3x - 2) + 1
y - 1 = 2(3x - 2)
(y - 1)/2 = 3x - 2
(y - 1)/2 + 2 = 3x
((y - 1)/2 + 2)/3 = x

That's exactly the unsimplified form of g^-1f^-1(x)! Notice that in solving, I undid one operation at a time, starting on the "outside": I subtracted 1, divide by 2, added 2, and divided by 3. That is, I first performed the inverse of f, and then performed the inverse of g. Because f is "on the outside", I undid that first, so that f^-1 is now "on the inside".

Another meaning of the term "invert" is to "turn inside out". That's what's happening!

Thanks so much, I'm somewhat clearer now.

Here's the relevant page from the book dealing with functions with the question we are dealing with (no. 13 d/e). It's a classroom book, as I said, so I guess they expect a teacher to fill in missing explanations. The explanations are often too brief for self tuition, in my view ( unless it's me being rather obtuse) with questions that get harder quite quickly.

 

[Dr.Peterson]

Thanks so much, I'm somewhat clearer now.

Here's the relevant page from the book dealing with functions with the question we are dealing with (no. 13 d/e). It's a classroom book, as I said, so I guess they expect a teacher to fill in missing explanations. The explanations are often too brief for self tuition, in my view ( unless it's me being rather obtuse) with questions that get harder quite quickly.

Clearly my original guess was right, that they just gave you these examples to introduce you to the idea that (fg)^-1(x) = g^-1f^-1(x), which they have not mentioned before. My discussions of the idea are the sort of thing you might have been expected to include in your comments for part (f), though at a much higher level than you would have. You are not expected to know these things at all.

It looks like this is just the second section in a chapter that introduces functions; they are going very quickly if they cover everything from the definition of basic function notation to rather complicated composites and inverses (though the algebra is not complicated -- all the functions are simple). It is possible that they do expect students to have seen the basics before.

 

[Simonsky]

Clearly my original guess was right, that they just gave you these examples to introduce you to the idea that (fg)^-1(x) = g^-1f^-1(x), which they have not mentioned before. My discussions of the idea are the sort of thing you might have been expected to include in your comments for part (f), though at a much higher level than you would have. You are not expected to know these things at all.

It looks like this is just the second section in a chapter that introduces functions; they are going very quickly if they cover everything from the definition of basic function notation to rather complicated composites and inverses (though the algebra is not complicated -- all the functions are simple). It is possible that they do expect students to have seen the basics before.

Indeed. I have found, with this book ( I'm about a quarter of the way through it) that more complicated concepts seem to suddenly 'jump out' at you from nowhere without adequate preparation -but I suppose these books aren't designed for self-study, so I shouldn't complain too much!

In fact the page I just posted was the introduction to functions, there was no related material before (except +,-,x, division of fractions and simplification).