Thanks in advance for any help. I think I might just be missing something simple.
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 .
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
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).
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.
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:
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.
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.