Thanks in advance for any help. I think I might just be missing something simple.
Problem:
Let f be a continuous function who's domain is the closed interval [0,1] and whose co-domain is also the closed interval [0,1]. Prove that there exists some value c in [0,1] such that f(c)=c .
Hints Provided:
1) consider the function g(x)=x-f(x) . Argue that g is continuous on [0,1] . Draw some inference about g(0) and g(1), specifically, are they positive or negative. Apply Intermediate Value Theorem to the function g.
What I have so far:
F(x) must be equal to x, ie. f(x)=x
0<=c<=1
0<=f(c)<=1
f(0)<=f(c)<=f(1)
Using the above I have proved that both f(c) and c are in the interval [0,1], but not that f(c)=c.
I kind of feel like I am not supposed to assume an actual function for f (not assume that f(x)=x), but I am not sure how to proceed then.
I'm not quite sure where the hint regarding g(x) fits in. If f(x) is in fact x then g(x)=x-x = 0, so neither positive or negative.
Thanks for the quick response!
I get that
but not why
Is this right?:
Using the theorem, I come up with a value of N that is equal to 0 because:Originally Posted by Intermediate Value Theorem
and because N is a value greater than and less than a value .
so such that .
So then c can be equal to any value [0,1] and g(c)=0 will still hold true, right? If so... that's the answer to the problem, I think.
You need to note that if f(0)= 0 we are done so we can assume that f(0) is NOT 0. And since f maps [0, 1] to [0, 1], we must have f(0)> 0 whence g(0)= 0- f(0)< 0.
If f(1)= 1 we are done so we can assume that f(1) is NOT 1 which means that f(1)< 1 and then g(1)= 1- f(1)> 0.
Since f is continuous, so is g and so there exist c such that g(c)= 0.
I'm... not sure I get it. We're assuming that even though we later prove that so ? Same for ?
Because the assignment was to prove that , why are we assuming the opposite?
Am I right in thinking we can assume what HallsofIvy stated because if , then
&
^
And that does not allow us to continue on to find
Ok... so, taking that into account, the function is only true along the domain (0,1), not [0,1], which means that becomes a piecewise defined function? Or that there's no way to know that (at the moment). But I did (like you said) successfully prove that for at least one , which was the assignment.
(Not a double post, separated intentionally as I am introducing a new question)
I'm not sure why I'm having such a hard time getting this
If I could trouble you for some further help:
This is part 2 of the problem, I tried to use what I got in part 1, but...
Part 2: Give an example to show that the result of Part 1 is not true if is not continuous, that is, give a example of a function whos domain is the closed interval [0,1] and whose co-domain is also the closed interval [0,1], but is not continuous and for any in [0,1].
For part of my answer of part 1, I argued that is continuous because becomes a polynomial (specifically a binomial), and polynomials are continuous at all points in their domain. I understand that if was not continuous, would not be, and therefore it would invalidate the rest of my work.
I'm having a hard time coming up with another with the specified domain (that is not continuous). If a function has a domain and range of [0,1], is there any function thats not continuous? I know that is not the only possible function (for example, would satisfy the requirements, but it is also continuous).
I don't think you really understand any of this.
Hall's suggestion is this: IF either then stop. You have found your .
So suppose that neither of those is true. We have , so that is clearly a continuous function.
Moreover, .
Thus so by the intermediate value theorem such that
But , therefore .
Oh! I get it. I'm not sure why I was hung up the way I was. Its not that for 0 and 1. You have to make that assumption to be able to prove that
Ok. Cool. Thanks!
emakerov, I read your post and went "oh, duh!" XD haha, thank you.
So I came up with the following:
so then:
Intermediate Value Theorem cannot be applied because is no longer a continuous function (even though all i did was mess with the one value).... and... therefore the result of part 1 ( ) is no longer true.
This does not satisfy the problem's requirements because the codomain of f is not [0, 1] (it's ).
Think about it in terms of graphs. You need to draw one or several lines inside the square [0, 1] x [0, 1] so that they don't intersect the diagonal y = x. You can shift around portions of these lines or even individual points, as long as the result is a graph of a function (i.e., there is one and only one point above every ). For example, can you make minimal changes to the graph of f(x) = 1/2 so that it does not intersect the diagonal?
Well, the graph of f(x)=1/2 is just a straight line on y=.5, but if you make f(x)=(1/2)x, then they don't intersect, because the slope is less. Except where f(0)=0. So just define it piecewise for that x, right?
But if we use the function f(x)=(1/2)x, then is the co-domain still [0,1]? With a domain of [0,1], f(x) will never be a value above .5 (though I suppose that would be the range, instead).
(f(x)=(1/2)x+1/2 also works, but with the same problem, just on 1).
^
But both of those functions are continuous, so I'd need something like
Plots on Wolfram Alpha
So then, most of the work transfers, that there is no way to get a using now that is not continuous.