# Continuity question, show that f(x) = 1 - x

1. Oct 28, 2009

### Nanatsu

1. The problem statement, all variables and given/known data

Suppose f (x) is a continuous function on [0;1], and 0 <= f(x) <= 1 for all x any [0;1].

(a)Show that f (x)= 1 - x for some number x.

(b)Prove the more general statement: Suppose g is continuous on [0,1] and g(0)= 1, g(1)= 0,then f(x)= g(x) for some number x.

2. Relevant equations
Unsure

3. The attempt at a solution

Neither makes sense to me. For the first one, isn't f(x) = x completely valid? Or a compressed sin function? Are they asking to show that 1-x is also valid, or that f(x) must be 1 - x no matter what?

For the second one, isn't a compressed cos function valid? It will be continuous on [0,1], g(0) = 1 and g(1) = 0. So why does g(x) = f(x) = 1-x only?

The question is copied word for word so tell me if I missed or misunderstood anything.

2. Oct 28, 2009

### arildno

This is complete nonsense.

Are you SURE you have given ALL relevant information here??

What book is this?

3. Oct 28, 2009

### union68

They're talking about the 1 X 1 square in the xy plane with line 1-x being one of its diagonals. If f is continuous in that square ( meaning 0 <= f(x) <= 1 for all x in [0,1]) then the function f must touch or cross the line 1-x at least once. Sure f(x)=x satisfies the question, but you must prove it for ANY continuous function f in that square.

Since they say that f is continuous on a closed interval and you're talking about a function crossing a line, I have a sneaking suspicion that the Intermediate Value Theorem is hiding in there somewhere.

Secondly, if g is a continuous function with g(0) = 1 and g(1) = 0, you must show that g intersects f somewhere. Draw a bunch of pictures depicting every case you can think of. You'll eventually convince yourself this must be true.

I've seen a problem like this before, so I had a head-start on deciphering it.

4. Oct 28, 2009

### Nanatsu

The question is from http://www.math.toronto.edu/~joel/137/assignments/a4.pdf [Broken]

Question 5 on the last page. I do believe I copied it word for word.

I think I understand what you're saying union. In other words I don't have to show that f(x) = 1-x, but that there must be at least one x where f(x) will equal a point of 1-x, or in other words f(x) must cross 1-x on some point?

Last edited by a moderator: May 4, 2017
5. Oct 28, 2009

### miqbal

You have to use the mean value theorem. The problem could be stated a bit more clearly by stating: show f(a) = 1 - a for some a $$\in$$ [0,1].

Since f(x) is continuous you can differentiate. Since 0 $$\leq$$ f(x) $$\leq$$ 1 on $$\left 0, 1\right$$ then then f'(x) will also be bounded on the interval.

The mean value theorem states that f'(z) = (f(b) - f(a))/(b - a) for some z in $$\left a, b\right$$. Should be straight forward.

6. Oct 28, 2009

### union68

The Mean Value Theorem does not apply since we are NOT allowed to assume that f is differentiable on (0,1). Remember, continuity does not imply differentiability...it is the converse that is true.

It is the Intermediate Value Theorem that is needed. You're going to have to look at a couple of different cases, but luckily some of them are trivial.

You may assume that f(0) < 1 and f(1) > 0 and ignore the other cases. Why?

EDIT: This last part is erroneous. I have deleted it. Let me hammer away a little bit more and I'll get back.

Last edited: Oct 28, 2009
7. Oct 28, 2009

### miqbal

I sucked at analysis :(.

8. Oct 28, 2009

### union68

Ha! This problem tripped me up some. I thought I had it worked out in my head and then I sat down to type it up and realized I was dead wrong. I hate it when that happens...

You are correct though, you want to show that f crosses the line y=1-x at some point. Let's denote the line as L(x)=1-x.

To rephrase the goal, you want to show that there exists an x such that f(x) = L(x). Or, equivalently, you could show that there exists x such that f(x) - L(x) = 0.

Furthermore, if you assume f(0) < 1 and f(1) > 0, then what can be said about f(0) - L(0) and f(1) - L(1)? Why can I throw away the cases f(0) = 1 and f(1) = 0? Will knowing something about f(0) - L(0) and f(1) - L(1) help you find a value of x such that f(x) - L(x) = 0? What can be said about the continuity of F - L?