1. I am having trouble with another math problem concerning inverses. My book says the following:

"

Basic Rule for Finding Inverses To find the inverse of a function f, we solve the equation

(f · f <sup>-1</sup>)(t) = t

for the function f <sup>-1</sup>(t)

EXAMPLE 1.40
Find the inverse of the function f(s) = 3s.

SOLUTION
We solve the equation

(f · f <sup>-1</sup>)(t) = t

This is the same as

f ( f <sup>-1</sup>(t)) = t

We can rewrite the last line as

3 · f <sup>-1</sup>(t) = t

or

f <sup>-1</sup>)(t) = t/3

"

Where I am confused is where the equation replaces (f ( f <sup>-1</sup>)(t)) for 3 · f <sup>-1</sup>(t).

I think I may have a very weak grasp of how it might work, but not any kind of solid rule to go by. This is what I've come up with: The value of a function for any given variable should be divisable by both the variable and a scaler that is not necesarily the same number for every variable value... What good is that knowledge though if the scaler is not always the same number? Perhaps the writer just recognized a pattern with this specific problem and used it to quickly solve the problem.

2.

3. I suggest you review the meaning of the term "function". Think of it as a machine into which you feed a number and get out another number (this may be the same as you input number).

Alternatively think of it as a rule that converts one number into another number.

In the present case, the "rule" f is "multiply by 3" and the input is any t. Any use to you?

4. Yes. I think I just have a weak grasp of the ideas of a function at the moment. What is a good analogy for the relationships between a function, its domain, its range, the codomain, and the notation used to describe these relationships? Third grade teachers use the old aligator eating the bigger number analogy for greater than or less than. Maybe we could make a three headed dog monster that would.... do something.

Bedtime is fast approaching. Maybe I'll dream of the codomain monster and wake up with a brilliant idea.

5. Your problem isn't with the domain, range, and codomain of a function. It's with what function notation means. In the example where f(s) = 3s, you're confused as to how they arrive at the equality:

f(f<sup>-1</sup>(t)) = 3f<sup>-1</sup>(t)

Try this on for size... let s = f<sup>-1</sup>(t). Then what is f(s)? From this, can you show the above equality?

6. I understand it now. It just seems to take me some extra thought for something that seems like it should be simple. It will probably just take a little time for me to be comfortable with it.

A trick question! For the function f<sup> -1</sup>(t) = s there are no values of t to apply values to, so the function has no inverse. Is this correct? Hmm... on second thought I think f<sup> 1</sup>(t) = s

I believe the inverse of f<sup> -1</sup>(s) = s would be f<sup> 1</sup>(s) = s

and the inverse of f<sup> 1</sup>(s) = 5s + 3 would be f<sup> -1</sup>(s) = (s - 3)/5

and the inverse of f<sup> 1</sup>(s) = s<sup> 2</sup> does not exist since there would be two range values for some values of s in the inverse.

7. Originally Posted by Demen Tolden
A trick question! For the function f<sup> -1</sup>(t) = s there are no values of t to apply values to, so the function has no inverse. Is this correct? Hmm... on second thought I think f<sup> 1</sup>(t) = s
OK, a coupla points: you don't need to write f<sup>1</sup> as this is just f!

Second, for f<sup>-1</sup>(t), s is in the codomain, so there need not be any t to apply f to. You want an s such that f(s) =t. Recall the definition of an inverse: f<sup>-1</sup>(f(x)) = x

I believe the inverse of f<sup> -1</sup>(s) = s would be f<sup> 1</sup>(s) = s
Now think this one through. Give f a rule, an easy one, what rule might f be such that f(x) = x = f<sup>-1</sup>(x) for all x?

and the inverse of f<sup> 1</sup>(s) = s<sup> 2</sup> does not exist since there would be two range values for some values of s in the inverse.
This much at least is true. Sooner or later you learn about injective, surjective and bijective functions, and you came close, but now is not the time for you.

8. Sorry, I got even lazier with my f(copy) = paste. Where the domain is copy, the range is paste and the codomain is T, which contains all typable text. Does the domain have to exists within the codomain?

Second, for f<sup> -1</sup>(t), s is in the codomain, so there need not be any t to apply f to. You want an s such that f(s) =t. Recall the definition of an inverse: f<sup> -1</sup>(f(x)) = x
Okay. I understand that s is in the codomain because s is in the range, or I think you could say the range of f<sup> -1</sup>(t) is [s,s], and the range is always within the codomain. I think I understand the relationship between f<sup> -1</sup>(t) and s, but I dont very well understand the relationship between s and t, except that there is a function that takes in t and spits out s. Are they really the same value?

Now think this one through. Give f a rule, an easy one, what rule might f be such that f(x) = x = f-1(x) for all x?
That the function multiplies the domain by 1.

9. If I write

f : s --> T

Then f is the function, s is the domain, and T is the codomain correct? Could I also write f(s) = 2x in the codomain of T? If this is true than 2x is really the definition of the function, or how we turn a number from the domain into a number in the codomain, and x is a number we grab out of the domain. If this is correct then the answer to one of my questions would be no, the domain and codomain are two different sets of numbers. Although they may contain some of the same numbers, this is not necesarily true.

AND the codomain may contain more numbers that may translate from the domain, and vice versa. I know we have already discussed a lot of this. It's just now that I am getting a good picture in my head of what I think the relationship is between all these ideas.

10. Crossed posts, but never mind for now.
Originally Posted by Demen Tolden
Does the domain have to exists within the codomain?Not at all, although at elementary level the domain and codomain are usually both R

Okay. I understand that s is in the codomain because s is in the range,
This doesn't seem quite right. A function is defined on its domain and codomain, it's an unarguable definition. The range is found. Do you see the difference?
the range of f<sup> -1</sup>(t) is [s,s],
tsk, tsk. There's something you're not quite getting, and I can't put my finger on. But let me say first that [s,s] makes no sense at all.

Anyway, let's try this. A gadget f: R ---> R (as yet unspecified) will be a well-behaved function if and only if

f is able to eat any element x in R and spit out one and only one element f(x) in R. Use that as a definition.

Note that this need not imply in general that for every y in the codomain there is some f(x) = y. Neither does it imply that for every y = f(x) there may not also be some y = f(w). But, where this implication does hold we will say that f has an inverse such that f((f<sup>-1</sup>)(x)) = f<sup>-1</sup>((f(x)) = x.

Let's return to your "range = [s,s]" stuff. For reasons you don't really need to know, the set R of real numbers is, in algebra, conveniently thought of as a "continuous" line, the Real Line. This line is by convention thought of as being the set-theoretic union of all open intervals of the form (a,b).

What's "set-theoretic union"? Nothing special, just that, given all numbers strictly greater that a and less than d, (a,d), and those similarly found in (b,e), the union of (a,d) and (b,e) is (a,e) where the common element c occurs only once.

So why use the union of open intervals to construct the real line? Because the real line has no beginning and no end, so the symbol ∞, which signifies this fact, must always be an "element" in some interval (a, ∞)

So, by agreement, all elements in the range of some f: R ---> R are found in some interval (a,b)

That the function multiplies the domain by 1.
So 1(x) = 1/1(x)? Looks good to me. Not a very interesting function, though!

11. Originally Posted by Demen Tolden
If I write

f : s --> T
Is this a typo? It is usual to use caps for sets. You meant S for the domain, right?

Could I also write f(s) = 2x in the codomain of T?
Hmm...if s is in S then f(s) is in T, for sure. The codomain of f is T, the phrase "codomain of T" makes no sense; T is the codomain of f.
the domain and codomain are two different sets of numbers. Although they may contain some of the same numbers, this is not necessarily true.
Again, hmmm... at the level we are talking, R is both domain and codomain, so they are the same sets, containing the same numbers.. But you're right; I can find a nice function from S to T where S and T have no elements in common, yet are still related via the function f. I advise you not to go there just yet, though.

12. My question was not a trick question. It was pretty straightforward. If f(s) = 3s, then f(whatever) = 3whatever. That's what function notation means. So f(f<sup>-1</sup>(t)) = 3f<sup>-1</sup>(t), which was the answer to your original question of how the author got from t = f(f<sup>-1</sup>(t)) to t = 3f<sup>-1</sup>(t).

You did, in the reply to my first post, correctly solve a couple of inverse problems. Kudos to you! But before we even learn all of the "-ective" definitions and even "codomain", we can come up with suitable "inverses" to functions which shouldn't at first glance have them. For example, f(x) = x^2 isn't invertible on the whole real line, as all positive numbers have two numbers mapped to them. But if we restrict our domain to the nonnegative reals, we have an invertible function. We can do the same for functions which have infinitely many points mapped to the same number, for example inverting trigonometric functions. We just restrict the domain of the function (for example, restrict sin(x) to -pi/2 to pi/2) and then invert it on that domain (so we get arcsin(x) defined from -1 to 1).

13. f : s --> T
I changed it to lowercase so i could write
f(s) = 2x
I think I made a mistake. I had originally meant that the s in f(s) is the same as the domain S, but since my understanding was still developing while I was writting my post, and at the end of it I began to believe that in something such as f(x) = 2x, x is a number taken from the domain and is not the domain itself.

Hmm...if s is in S then f(s) is in T, for sure. The codomain of f is T, the phrase "codomain of T" makes no sense; T is the codomain of f.
I understand why now I think.

Again, hmmm... at the level we are talking, R is both domain and codomain, so they are the same sets, containing the same numbers.. But you're right; I can find a nice function from S to T where S and T have no elements in common, yet are still related via the function f. I advise you not to go there just yet, though.
It's not too much of a problem for me, but it is just a little bit more confusing to take from one number pool a number, convert it, and put it back into another number pool that has the same numbers. I guess its not so bad though.

.................................................. ..........

A function is defined on its domain and codomain, it's an unarguable definition. The range is found. Do you see the difference?
I think I do. Then the range should be (-,). Is f(x) = s just useless knowledge then since it doesnt say what to do to the value you take from the domain? Unless of course you find a way to figure out what s is.

Note that this need not imply in general that for every y in the codomain there is some f(x) = y. Neither does it imply that for every y = f(x) there may not also be some y = f(w). But, where this implication does hold we will say that f has an inverse such that f((f-1)(x)) = f-1((f(x)) = x.
What would be the difference between x and w? The inverse funciton then is the function that you would use to retrieve the domain value x using a codomain value f(x), correct?

So I have more concepts than just the domain, codomain, function, and range. Now there is a domain value such as x in f(x) and the range value which is f(x).

14. Originally Posted by Demen Tolden
f(x) = 2x, x is a number taken from the domain and is not the domain itself.
Correct, good.

confusing to take from one number pool a number, convert it, and put it back into another number pool that has the same numbers. I guess its not so bad though.
Confusing? Let f(x) = 2x, let x in R = 2. What is f(2) and what "pool" is it in?

Is f(x) = s just useless knowledge then, since it doesnt say what to do to the value you take from the domain? Unless of course you find a way to figure out what s is.
No no no, wrong way round!! You always know that x is an element in the domain, you always know that s is an element in the codomain ; f tells you how to find s from x, as in f(x) = s, that's what the notation means.

The inverse function then is the function that you would use to retrieve the domain value x using a codomain value f(x), correct?
Bingo!! Now try, using the information you have in hand, to come up with the circumstances where this might be possible.

Here's a hint: in general, we don't assume that for every y in the codomain there is some f(x) = y, and we don't assume that, in general, for every f(x) there is a unique x. Now what do you find if you do make both these assumptions (not that they really are assumptions)

Try for maximum precision (unlike me, some might say!)

15. Confusing? Let f(x) = 2x, let x in R = 2. What is f(2) and what "pool" is it in?
f(2), which is a variable in the range, which exists in the codomain has the value of 4. 2 is the value of the variable taken out of the domain and entered into the function f. The range of f is (-∞, ∞).

No no no, wrong way round!! You always know that x is an element in the domain, you always know that s is an element in the codomain ; f tells you how to find s from x, as in f(x) = s, that's what the notation means.
I think I understand the meaning of the notation now, but what I am saying is that I don't see what to do with a domain value to be able to find a range value. Should I imagine the function says f(x) = s + x<sup>0</sup> so that if I need to find the inverse I would get s +( f<sup> -1</sup>(x))<sup>0</sup> = s... no that doesn't work out. With what I know now, I don't see how f(x) = s is a viable function.

Here's a hint: in general, we don't assume that for every y in the codomain there is some f(x) = y, and we don't assume that, in general, for every f(x) there is a unique x. Now what do you find if you do make both these assumptions
Every codomain value (y) can be attained through a function (f) of a particular domain value (x) unique to the codomain value. In other words f(x) = y. How about that. But then... where is the function? What is the funciton that translates the two? Is it just a mystery? Does f really represent a variable that represents a function?

16. Should I imagine the function says f(x) = s + x<sup>0</sup> so that if I need to find the inverse I would get s +( f -1(x))<sup>0</sup> = s... no that doesn't work out. With what I know now, I don't see how f(x) = s is a viable function.
Oops, I see a mistake. How about 0x instead of x<sup>0</sup>. Would the inverse of f(x) = s be s = s? But that doesn't give you a domain value does it?

17. Originally Posted by Demen Tolden
Every codomain value (y) can be attained through a function (f) of a particular domain value (x) unique to the codomain value. In other words f(x) = y. How about that.
OK, there's your problem right there. Don't try to go any further until we've fixed this. Read this bit very carefully, as I am choosing my words with great care:

Let S and T be sets, and let f be a function f: S --->T. For each s in S there is some t in T such that f(s) = t. Let g: T ---> S, such that for each t in T there is some s in S where g(t) = s..

Using this information only, tell me a) whether I may assume that f and g are mutual inverses and b) if not, why not; what additional properties of f (or g) I might need to ensure that they are.

EDIT: Maybe that's a bit unfair. Try this instead (or first). Suppose f is a function on N, the counting numbers 1, 2, 3,.... Then N is clearly the domain of f. When f(x) = 2x for all x in N, what is the codomain of f?
Does f really represent a variable that represents a function?
Check the definition of a function we gave. Hint: it's not a variable.

18. This conversation is really getting needlessly difficult. Demen Tolden, you learned calculus once, but you either learned it in notation that is inconsistent with "standard" notation (i.e. the notation that Guitarist is using), or you have forgotten not only the calculus but also the notation.

In any case, introducing mathematical abstraction is not the way to go about solving this dilemma. Instead, I think it makes sense to reintroduce the idea of functions as particular equations relating two real variables x and y, and then working in the function notation.

If you want to go down this road, let me know. I'm more than willing to help. If not, I don't want to be a spoilsport or naysayer, but I predict that this discussion will only continue to confound Demen Tolden.

19. Yeah, you're probably right. Let me in (partial) mitigation say that, although my primary objective was to help our friend through this particular blind spot (we all have them!), I was trying to frame my help in a way that would make more difficult, future, concepts seem more natural.

Not much use if it doesn't achieve the primary objective, though! So take it away!

20. Lets head down the mathamatical abstraction road a little while longer. Serpicojr, you are probably right about functional notation being a better road to go on first, but I feel I am close to graduating from the school of mathamatical abstraction grade one, and I wouldn't want to give up now.

Unless I really am still a long way off. I'll be working on my next post now.

21. I think Guitarist's last post is pretty good, as he really explains in clear mathematical terms what the statement "f(s) = t" means. I'll concede to your wishes for the time being--I never want to dissuade someone from learning about something in a manner that they enjoy! However, if it looks to me like you're still having trouble with the "f(s) = t" statement, I'll jump in. After rereading your last couple of posts, it looks like you are on the cusp of understanding this, and it looks like Guitarist's last post might give you the last little nudge for you to attain edification.

I'll be gone for the weekend, so you two will have time to work without me butting in.

22. First I want to do the definition of a function. A function is a rule that applies to a domain value to get a range value. From what I understand in this notation f: S --> T, f represents an undefined function. In this notation f(x) = 2x, f represents a function that says to take a number from the domain and multiply it by 2 to get a number in the range of f that has a relationship with x. But in something such as f(x) = s, I don't understand because it doesn't say what to do with a number you take from the domain as far as I can see.

Let S and T be sets, and let f be a function f: S --->T. For each s in S there is some t in T such that f(s) = t. Let g: T ---> S, such that for each t in T there is some s in S where g(t) = s..

Using this information only, tell me a) whether I may assume that f and g are mutual inverses and b) if not, why not; what additional properties of f (or g) I might need to ensure that they are.
a)not quite, but almost
b)f is a rule that applies to a domain value, and the inverse of f is also a rule that applies to a domain value, but if

f(s) = t

then you could write

g(f(s)) = s

If

f(f<sup> -1</sup>(s)) = s

then

f(f<sup> -1</sup>(s)) = g(f(s))

Hmm, I might have hit a roadblock.

EDIT: Maybe that's a bit unfair. Try this instead (or first). Suppose f is a function on N, the counting numbers 1, 2, 3,.... Then N is clearly the domain of f. When f(x) = 2x for all x in N, what is the codomain of f?
The codomain might be all even numbers greater than 0, or in notation (0, ∞)

23. Wait a minute, g would have to be the inverse of f. That is the only way you could retrieve a domain value from a range value, or the only way that you could enter in an s value in the function f and get t, then enter in that same t into the function g to get x again.

24. OK, Demen, I see now that serpicojr was right; I can't make you see it, so I'm not explaining this stuff well enough. Please wait for his return from his weekend away.

I feel I let you down, and I'm sorry.

25. Shucks, thanks for trying I guess. I was surprised you kept working with me on this for 18 hours.

26. The following is not instructional or informative, nor is it really meant to refresh. I'm assuming you're comfortable in and out with pre-college mathematics, and I'm just trying to emphasize a common thread amongst courses one encounters in secondary school.

Okay, so functions first arise in the context of an algebra class--by which I mean the year-long algebra class US students take during 8th or 9th grade (9th being the first year of high school). We encounter them first as graphs of equations. The simplest graph is that of the line--y = mx + b, where m is the slope, b is the y-intercept, and we're assuming the line is not vertical.

After a year of geometry, another year of algebra is taken, and some more equations are introduced--parabolas, ellipses, hyperbolas. We typically deal with parabolas which open upward or downward and have horizontal symmetry. These can always be given the form y = ax^2 + bx + c. This is much like what we have for lines--we have y on one side of our equation and some crap on the other side. Unlike the case with the lines, though, we can't solve for x in terms of y with just one equation. Hyperbolas are similar--we can consider them as given by equations like y = k/x, where k is some nonzero number. Ellipses can be treated similarly, but we have to be a little careful. (We also learn about higher degree polynomials and how we may go about graphing them, I think, but this doesn't have as nice a presentation as the quadratic/parabola case.)

Also in this year, we learn about the trigonometric ratios sine, cosine, and tangent. The important aspect of these ratios for our conversation is that, given an angle, we get a ratio.

The following year, we take a precalculus course. This course introduces concept such as the exponential and logarithm (actually, they may be presented in the second algebra course, I don't recall), and we learn about the graphs of equations like y = e^x and y = ln x. We also extend the trigonometric ratios to make sense for any real value, and we graph equations like y = sin x, y = cos x, y = tan x.

So the important thing to get out of this discussion is that, through many years of math, we are trained to think about equations of the form y = "something involving x", largely in the context of graphing them. (There's also the context of their practical applications and their mathematical meaning, but graphs are really a good way to think about functions, especially for calculus.)

Let me know if you need any clarification about what I'm getting at. If not, I'll proceed.

27. You are clear to procede sir. If it will help you any to explain, I'll give you a little bit more about my math related background. I took all the math classes I could when I was in highschool, and took AP Calc in my senior year, but that year I was really having a much stronger developing interest in art and ended up going to college for computer animation. Computer animation may relate to this because after working in it for a long time I've developed a stronger "3D imagination." I am 28 years old now, and haven't used much math beyond algebra since my highschool years. I hope this information might help you to help me.

28. Your background info is very useful--thanks for providing it!

So the key points that I wanted to make in the last post were that graphs can represent equations that relate two variables x and y, and oftentimes we can write these equations in the form y = "some equation involving x and no other variables".

The idea of a function arises here. Graphs of equations of the form "y = 'something involving x'" satisfy a special property--for each value of x, there is at most one value of y (so that the point (x,y) is on the graph). This is the same as the vertical line test: if we draw a vertical line through such a graph, we hit at most one point. Examples of graphs which satisfy these equations were given last time:

1. Lines: y = mx+b
2. Parabolas: y = ax^2 + bx + c
3. Hyperbolas: y = k/x
4. Trig functions: e.g. y = sin x
5. Exponentials and logs: y = e^x, y = ln x

A nonexample of such a graph is that of an ellipse: (x/a)^2 + (y/b)^2 = 1. When y = 0, x can be a or -a. (However, we can salvage things in this case--we can break this graph into its top half, x = a sqrt(1-(y/b)^2), and bottom half, x = -a sqrt(1-(y/b)^2).)

We can now introduce a function in a couple of ways:

1. It is a graph satisfying the vertical line test.

2. It is a rule associating at most one value y to each value x.

3. It is the right hand side of an equation y = "something involving x".

The second and third descriptions will now become more important to us, but it's important to keep in mind a graphical representation of functions as associating one variable to another.

I've got to run, so I'll pick this up later.

29. Okay. I follow you so far.

30. So let's think about functions as rules associating a variable x to a variable y or as the right side of an equation of the form y = "some formula in x". In this setting, we may be interested in repeatedly applying a function, or we may be interested in performing one function and then another. A graph can help us to this end somewhat, but things will get pretty messy pretty quickly. And explicitly writing out the formula that gives the function can be cumbersome. So we introduce function notation.

When we write:

f(x) = some expression in x

we mean that, no matter what we put into the parenthesis, we replace all instances of x with that thing on the right. So, for example, if we have:

f(x) = x^2

then we have:

f(2) = 2^2 = 4

and:

f(-1) = (-1)^2 = 1

and even:

f(dog) = dog^2

This notation really emphasizes the rule-like aspect of functions: you input something, and the function transforms it according to some rule. And this also emphasizes the equation aspect--the function is, in this case and in most cases you're concerned with in calculus, given by some equation.

Let me take a moment to note that we've already encountered function notation. For example, we've seen sin(x), cos(x), tan(x), and ln(x). These are all functions which take a value x and transform it according to some rule. We don't really have an equation for these guys in terms of x--they're described by some sort of geometric or algebraic process. But we use these functions so much that we given them distinguished names.

But going back to repeated iterations of a function and applications of different functions. Let f(x) = x^2. If I wanted to apply this function twice, function notation allows me to write f(f(x))--that is, I apply f to x once (this is the inner f(x)), and then I apply f to the result. The sum total of this effect is:

f(f(x))
= f(x^2) [This is applying the inner f to x and getting x^2.]
= (x^2)^2 [This is applying f to x^2.]
= x^4

If I were to just write my function as y = x^2, I wouldn't have this convenient notation. This is kind of a simple example, but imagine what a bear it would be to consider how to write down what you get when you the function y = sqrt(tan(x)-4x^2+9)/e^(2x+sin(ln(x)) twice. If we wrote this function in function notation, we'd sweep things under the rug by simply writing f(f(x)). Function notation is a notational convenience--it doesn't necessarily help us calculate things.

Note above that I had some functions sitting inside some other functions--for example, sin(ln(x)). Function notation makes it easy to express this in general--if I have two functions f and g, and I want to apply g to x and then f to the result, I write f(g(x)). We also write this as f⋅g(x) (really, this should be a small open circle between f and g, but this is the best I could find), and we say this as "f composed with g". So now we have a nice, easy way of writing new functions in terms of old functions. Consider the example:

f(x) = sin(x)

g(x) = 1 - x^2

h(x) = sqrt(x) (assuming x ≥ 0)

Then if I consider h⋅g⋅f(x), I get:

h⋅g⋅f(x) = h(g(f(x))) = h(g(sin(x)) = h(1 - (sin(x))^2)

I use the identity 1 - (sin(x))^2 = (cos(x))^2 and get:

= h((cos(x)^2) = cos(x)

Nice.

Main points here: function notation emphasizes the rule and equation aspects of functions, allows us to name special functions and makes it easier to deal with a lot of functions at once and repeated application of functions. Cool?

31. Yup. Cool

32. Alright, so you clearly understand the motivation for functions and function notation. So here's the kicker. I can think of functions as graphs:

y = "formula in x"

or as rules:

f(x) = "formula in x"

so f(x) and y look like they play the same role. This is a fine interpretation. But in some sense they play different roles. f(x) is some sort of formula or rule which acts on the variable x; y is a variable which is depending on the variable x. y may vary depending on x by some rule or formula f(x), and in this case we might write:

y = f(x)

So let's think about what this is saying. Literally, this is saying that x and y are variables, f is some rule or formula acting on x, and y depends on x by being equal to the formula f applied to x. If I have a value for x, I can find the corresponding value for y by plugging it into x. This is not saying that any value x gets sent to the same value y. This is not saying that y is a function.

For example, think of y being velocity of some object and x being time. If I write y = f(x), I mean that the velocity of this object happens to depend on time according to some rule f. x and y are just quantities I can measure, and the equation just tells me that if I know x, I can find y.

So if I'm given some function (in the sense we're talking about now) y = f(x) and I'm asked to find the inverse, I'm really saying... can I find a function g which satisfies:

1. g is a function with INDEPENDENT variable y and DEPENDENT variable x, i.e. x = g(y) (this is not a formula, just a relationship between x and y via g);

2. g undoes what f does--when I compose g with f, I get g(f(x)) = x (this is a formula here--when I plug x into gf, I just get back x)--and f undoes what g does--when I compose f with g, I get f(g(y)) = y (note I use the variable y here because g acts on that variable).

Then what your book was saying is this... if we know what f is (i.e., we have a formula for f) and want to figure out what g is, we may be able to use the relation f(g(y)) = y to find a formula for g. We just write out what f does to g(y) and then solve for y. For example...

f(x) = ln (x-4)

Then:

y = f(g(y)) = ln(g(y)-4)

e^y = g(y)-4

g(y) = e^y+4

(Also, your book just happens to be using s instead of x and t instead of y.)

Does this make sense? If so, you may proceed with Guitarist and talk about codomains and whatnot.

33. Well, after this excellently lucid exposition, I'm going to lower the tone significantly.

Consider a restaurant menu. Notice first that you have two sorts of entries: meals and money. Next you notice that to each meal you can associate a sum of money, the "price", and no meal has no associated price. Then notice that no meal associates to more than one sum of money. Next notice that several meals associate to the same price. Finally notice that there is no meal costing say exactly 10 euros.

Plonk down 20 euros and ask the waiter to bring you the corresponding meal. What will he bring today? What will he bring tomorrow? You have no way of knowing, since there are several meals priced at 20 euros. In other words, knowing that you have 20 euros gives you no knowledge about the meal you're getting

Lets try and put this in something resembling mathematical language. To each element in the set Meals there is some element in the set Money and the two sets are connected by the function "price of". You may say that y (euros) = price of x meal, or better, y = function of x. The term "function of" is abbreviated to "f of" and is written f(x). By my example above, you cannot find an inverse function that takes y = 20 and always returns x = lasagne, neither can you find a function that takes 1 euro and returns anything at all.

We say that f is invertible iff for every y in Y there is some x in X such that y = f(x) and if f(x<sub>1</sub>) = f(x<sub>2</sub>) in Y this implies x<sub>1</sub> = x<sub>2</sub> in X.

Can you see how this definition arises out of my parable above?

34. Sorry I took so long to post, I was reading my new calc book. I think I understand most of this function stuff now. I think I still have a few rusty areas with functions in logrithims and trig. I'm sure these will just take practice. My trouble with logrithims arises from me never having practice in it, and trig from me having not used it for 10 years. Trig can probably just be summarized with the pythagorean theorm and soh cah toa. I just need to see the patterns I used to.

Unless you have a particular train of thought running or problem you think I should figure out, I am having trouble with finding the inverse of this problem in particular:

f(x) = 3 + x<sup>2</sup> + tan(πx/2) where -1<x<1

I've stared at it for 2 hours, tried a few things, but havnt figured it out. What I would like to do is replace f(x) with y, get x alone on one side, and then interchange x and y, replace y with f<sup> -1</sup>(x) and bingo, there is the inverse.

Thats a bad looking pie too (π).

35. Do you know if this question asks you to find the inverse or simply asks you to say whether it exists or not?

36. Yes, I know it may seem quite a step up from what we've been doing, but I'm ready for it I think.

The question actually says find f<sup> -1</sup>(3)

and then after that find f(f<sup> -1</sup>(5))

Some of the latest things I have been working on in my new book are inverse functions dealing with trig functions.

37. Wait a minute. All I have to do is this:

since f(x) = y

then f<sup> -1</sup>(y) = x

if i have f(x) = 3 + x<sup>2</sup> + tan(πx/2)

then all I have to do is plug in 3 for f(x) and I get

3 = 3 + x<sup>2</sup> + tan(πx/2)

0 = x<sup>2</sup> + tan(πx/2)

x<sup>2</sup> = -tan(πx/2)

Hmm... where to go from here...

Maybe the next step has to do with changing tan to arctan

38. Well, as the numerical answers to both these problems are completely trivial, I suspect you are merely being asked to repeat the properties of functions and their inverses, from which the answers follow immediately.. I doubt you are being asked to show that this function is invertible, but I may be wrong.

For f(x) = y, then f<sup>-1</sup>(y) = x, you won! Actually for f(x) = 3, you had the answer in your hand, but then dropped it. Look again.

The answer to f(f<sup>-1</sup>(5)) is equally trivial once you write down the other property you're being asked to remember.

You do know it, because you told me about it a few days ago.

39. I'm sorry I don't know what you mean about f<sup> -1</sup>(3). I think in order for me to find the answer I need to first figure out the inverse of f(x), and I don't know that I can find the inverse of f(x) if I can't get the x out of the tangent angle. I know f(f<sup> -1</sup>(5)) isn't so tough since the functions cancel eachother out.

I'm learning. But this inverse trig stuff is frustrating me.

40. You reduced finding f<sup>-1</sup>(3) to finding all x between -1 and 1 which satisfy:

x^2 = -tan(πx/2)

Try graphing y = x^2 and y = -tan(πx/2) out and seeing where they intersect. There is one "obvious" solution, and the problem suggests that it's the only one.

41. 0? That's too easy. I kind of wanted to find it by reasoning it out. I kind of feel that graphing is cheating when I don't know something very well.

Oh well, that's the answer, and I'll move on.

42. Using the tools that are available to you is not cheating. You're willing to learn math from a book rather than derive it all yourself; you should be willing to use a calculator to graph things if you're not yet comfortable with doing so yourself. There's no shame to be had.

The only way I can think of to show that no other solutions exist requires knowing that tan x > x for 0 < x < π/2. I'm sure there's a geometric argument, but it's not immediate to me. And there's a calculus argument (which really depends on the geometric argument above), but you don't have that under your belt. If you come up with something, I'd love to see it.

43. Yea well I guess I'll go with graphing every once in a while.

I should know this one but I have worked out the theory of reletivity to be an inverse, but I need to figure out what practical application it has.

m = f(v) = m<sub>0</sub>/(sqrt(1- v<sup>2</sup>/c<sup>2</sup>))

the inverse is

f<sup> -1</sup>(v) = sqrt(c<sup>2</sup> - c<sup>2</sup>v<sup>2</sup>m<sub>0</sub><sup>2</sup>)

I made this diagram of a clockwise cycle with the domain variable cycling through the funcion f machine to turn into a range variable, then the range variable cycleing clockwise through an inverse function f<sup> -1</sup> machine to turn back into a domain variable. In this case of the theory of reletivity the domain should be v, the velocity of a particle, and the domain should be m, the reletivistic mass of a particle. Does the knowledge of f<sup> -1</sup> just allow us to travel from the domain variable v counterclockwise through the inverse function to get the range variable m?

----------------------------------- > -[ f ]- >
------------------------------ > ----------------- >
-------------------------(v)---------------------------(m)
------------------------------ < ------------------ <
----------------------------------- < -[ f<sup> -1</sup> ]- <

44. Seems I may not understand function basics yet.

In order for f<sup> -1</sup> to be useful shouldn't it be in terms of the range variable, m, in the case of the theory of reletivity? Then if you knew the mass you could find the velocity.

So instead of this as the inverse function:

f<sup> -1</sup>(v) = sqrt(c<sup>2</sup> - c<sup>2</sup>v<sup>2</sup>m<sub>0</sub><sup>2</sup>)

This might make more sense:

f<sup> -1</sup>(m) = sqrt(c<sup>2</sup> - c<sup>2</sup>m<sup>2</sup>m<sub>0</sub><sup>2</sup>)

and would allow you to discern the velocity of a particle if you knew the reletivistic mass.

To get the original inverse I used this process:
replace f(x) with y, get x alone on one side, and then interchange x and y, replace y with f<sup> -1</sup>(x) and bingo, there is the inverse.
But in terms of x and y, wouldn't it make more sense to not interchange x and y, but instead replace y with f<sup> -1</sup>(y)?

45. Your second post is correct--if we think of a function f as giving a dependent variable m as a function of an independent variable v, the inverse function f<sup>-1</sup> should give v as a function of m. (Well, mathematically speaking, we can use whatever letters we want, so long as we make it clear what they mean; but for practical purposes, e.g. when variables represent some measured quantity, it makes sense to give them names and stick with them.)

Be careful with your algebra there... I think you should have (m<sub>0</sub>/m)<sup>2</sup>. And you may as well pull the c<sup>2</sup> out of the square root as c.

46. Oops, you are right. Sorry about that.

I've got another one.

How would you "Find a formula for the inverse of the function:"

f(x) = (4x-1)/(2x+3)

This is some of my work on it:

y = (4x-1)/(2x+3)

Then I thought about maybe useing the natural log to seperate the x's

ln y = ln (4x-1) - ln (2x+3)

but that doesn't seem to go anywhere unless there is some property of logs that I am missing or don't know.

The graph of the function seems to be a hyperbola with a verticle asymptope at about x = -1.5 and a horizontal asymptope at y = 2. I believe that means that the inverse should have a verticle asymptope at x=2 and a horizontal asymptope at y = -1.5. Still, this information isn't what I need to solve the problem.

47. You're making this too hard on yourself. You eventually want to find x = g(y) for some function g. With your work, we'd eventually have to take exponentials to get x out of the logs. But this is really just undoing whatever work you already did, so you may as well have not taken the logs in the first place. Instead, I think there's a very simple idea which just begs to be used--cross-multiplication.

48. Oh, I see it.

f<sup> -1</sup>(y)=(3y+1)/(4-2y)

49. Hmm, your denominator doesn't look quite right, though I haven't bothered to work it out. I suspect you should switch your coefficients?

Anyway, here is the general rule for finding the inverse of a function:

Let f(x) = {something in x}

Set f(x) = y, and solve for x in terms of y (if possible)

Then replace every occurrence of y by by x, which should yield f<sup>-1</sup>(x).

And what if you can't solve for x in terms of y? Graph it!

50. This might sound odd. I know I am supposed to switch x and y and have the inverse start with f<sup> -1</sup>(x), but it seems to make more sense to me if I leave y in the equation and start with f<sup> -1</sup>(y). Where x is the independent variable, and y is the dependent variable, it doesn't make sense to me to write the inverse function "of (x)" when you are using y to figure out what x is. I guess the definition of the domains and variables are supposed to get swapped when the inverse is considered rather than writing the inverse in terms of the dependent variable.

I've got another problem too. As far as I know C is not a constant, but I could be wrong.

Sove this equation for x:

e<sup>ax</sup> = Ce<sup>bx</sup>, where a does not equal b

I've broken the equation down to ln (x/x<sup>C</sup>) = ln (b<sup>C</sup>/a)

If C where a constant, I would say that would be good enough.

51. It's all a matter of taste. If you want to think of x and y as variables related to each other via a function f from x to y, then, sure, think of the inverse of f as going from y to x. If you want to keep the convention that x always represents the independent variable and y the dependent, then you have to swap x and y when finding an inverse. But it's really no big deal--x and y are just symbols, and we can make them stand for whatever we want.

As far as your next problem, your answer is not correct (I think you need to be careful when taking logs), and you didn't phrase your answer in the way the question asked. The question says solve for x. That means write x = something in terms of a, b, and C.

x = (ln C)/(a-b)

I think I am pretty comfortable with everything except inverse trig now. A little more practice and I should be fine I'm sure.

53. This I think is the only type of inverse problem left that stumps me.

Prove that:

cos(sin<sup> -1</sup>x) = sqrt(1-x<sup>2</sup>)

My work on this problem:

Actualy i just figured out this problem by making a triangle diagram. I set sin<sup> -1</sup>x = Θ and then set the length of the side opposite Θ to x, and the hypotenuse to 1 acording to the inverse of sin<sup> -1</sup>x = Θ, which is sin Θ = x. The next step was to find cos Θ since cos(sin<sup> -1</sup>x) is the same as cos Θ, which finally gave me sqrt(1-x<sup>2</sup>).

54. Good job with both of your last posts. In particular, you should be proud of your solution to the last one--it shows understanding of the functions at hand and ingenuity in problem solving. That's actually a standard method for calculating such things that students have to learn for trigonometric substitutions in second semester calculus, and whereas many students struggle to use the method, you came up with it yourself!

55. Yeah, I suspected the inverse trig functions, which are not typical, would crop up sooner or later, that's why I went babbling about domains, codomains and ranges.

But first let me echo serpicojr's sentiments: Demen I am extremely impressed by your rapid progress here, well done.

Now, the trig functions and their inverses: first, being periodic, none of them is strictly invertible. To see this, consider the sine function. The domain of this function is all R, the range though is [-1, 1].; that is, for all Θ in R, sin Θ ∈ [-1, 1]. By the periodicity of the sine function, given x in [-1, 1], there is no unique Θ = sin<sup>-1</sup> x. In other words, given sin Θ = x and sin φ =y, if x = y I may not assume that Θ = φ.

But, also by the periodicity of the sine function, there is a "work-around".

Choose the smallest subset S of R such that, for all Θ<sub>s</sub> ∈ S, sin Θ<sub>s</sub> covers [-1, 1]. Under this restriction, the sine function is invertible.

Now, obviously, by the periodicity of the sine function, there are many such S. By convention one chooses S so that it includes the zero of R, that is, S = [-π/2, π/2].

This is not quite as arbitrary as it seems; any other subset of R can be used under appropriate scaling.

56. I have 2 questions reguarding Guitarist's post. The first is about φ (varphi?). I've never really seen it before, and I'm sure it is probably just an alternative to useing Θ when you have already used Θ. My question is that typically when you use φ does φ usually have a relation to Θ for example like y is usually the dependent variable of x? My second question is about zero of R. What is meant by this?

Also I don't know that I should get so much credit for using a triangle diagram to solve the inverse trig function. I used that method because I couldn't figure the problem out using trig identities.

Nevermind about the zero of R. I understand this now.
So that the set S includes 0.

57. Originally Posted by Demen Tolden
I have 2 questions reguarding Guitarist's post. The first is about φ (varphi?). I've never really seen it before, and I'm sure it is probably just an alternative to using Θ when you have already used Θ.
Yes, right. It's another variable in the domain of sine. I guess I could have used Θ<sub>1</sub> and Θ<sub>2</sub>, but I chose not to. It's not a big deal.
My question is that typically when you use φ does φ usually have a relation to Θ
No, it's just an independent variable in R.
Also I don't know that I should get so much credit for using a triangle diagram to solve the inverse trig function. I used that method because I couldn't figure the problem out using trig identities.
Yes, but as your tutor said, you used the available information to solve the problem. No more is required of you. You did very well

So that the set S includes 0.
Yes

58. Sweet, so I finished Chapter 1:Functions and Models! Yay! /CHEER

Thanks for all your help guys! Unless you guys have any objections or think there's a few more concepts I should know before I move on, next up is:

CHAPTER 2: Limits and Derivitives.

I'll probably start a new thread for that.

59. Originally Posted by Demen Tolden
, next up is:

CHAPTER 2: Limits and Derivatives.

I'll probably start a new thread for that. :)
Yes do, but let me ask you this.

What book are you working from? It doesn't seem to me like the logical next step, unless you have already been told about series, sequences, convergence and the like. I may be wrong, but it seems to me you'll need these concepts in hand before you get into limits and derivatives.

Or maybe you were planning to start there?

60. I posted a good answer at this link so it could be used easily in the future:

http://www.thescienceforum.com/viewt...?p=89843#89843

I don't what you mean obviously, but I could look it up.

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