True or False- Functions and Continuity

1. Jun 14, 2013

Justabeginner

1. The problem statement, all variables and given/known data
True or False. If f(x) is continuous and 0≤ f(x) ≤ 1 for all x in the interval [0,1] then for some number, x, f(x)= x. Explain your answer.

2. Relevant equations

3. The attempt at a solution

False. I think that even though 0 and 1 are included in the domain, it is possible for the x and y values NOT to coincide.
(I googled zig zag function, because I didn't know the name, and I found the 'Weierstrass' function. I thought this would be an acceptable counterexample?)

Thank you.

2. Jun 14, 2013

dirk_mec1

Would this satisfy: f(x) = x+1 ?

3. Jun 14, 2013

Justabeginner

No it wouldn't. This is a one-to-one function.

4. Jun 14, 2013

Staff: Mentor

Justabeginner, your answer is correct, but the reason isn't. f(x) = x + 1 is one-to-one, but so what? This function doesn't satisfy the requirement that 0 ≤ f(x) ≤ 1 for all x in the interval [0, 1].

For any function that satisfies the problem requirements, is it possible for the graph to NOT intersect the line y = x?

Last edited: Jun 14, 2013
5. Jun 14, 2013

Office_Shredder

Staff Emeritus
Draw a picture! The function's graph lies in the box [0,1]x[0,1]. The line y=x cuts that box in half. The points (0,f(0)) and (1,f(1)) are located where on that box?

6. Jun 14, 2013

Justabeginner

They are the left lower and right upper vertices of the box.

And I think it is possible for the function to NOT intersect with the line y=x.

7. Jun 14, 2013

Office_Shredder

Staff Emeritus

No, those are the points (0,0) and (1,1), not (0,f(0)) and (1,f(1)) in general.

Also your two sentences are contradictory. If f(0) = 0, then the graph does intersect the line y=x and the origin

8. Jun 14, 2013

LCKurtz

@justabeginner I am curious to know whether these questions you are posting are from a class you are taking or from self studying.

9. Jun 14, 2013

Staff: Mentor

In another thread he (she?) said this was self-study. From the questions asked, he/she is doing differential and integral calculus at the same time, which is not a good idea, IMO.

10. Jun 14, 2013

Infrared

Hint: Consider the function g defined by $g(x)=f(x)-x$. Note that by the constraint $0 \leq f(x) \leq 1$, $0 \leq g(0) \leq 1$ and that $-1 \leq g(1) \leq 0$. How can you use this along with g's continuity to show that $g$ has a zero in $[0,1]$?

11. Jun 14, 2013

YawningDog27

Here's a hint: consider the function $g(x) = f(x) - x$ for $x \in [0,1]$. Is $g$ continuous, and what can you say about $g(0)$ and $g(1)$?

oops...sorry HS-Scientist for pretty much repeating your hint.

Last edited: Jun 14, 2013
12. Jun 14, 2013

LCKurtz

As is usual on PF, the hints escalate until soon the problem will be completely worked.

13. Jun 15, 2013

Justabeginner

(I was out on the road all day,so I am SO sorry about not responding to these posts till now).

LCKurtz- I'm self-teaching myself Calculus AB and BC, and am just using a textbook, and doing everything in order. The problems that I post are the problems I have difficulty with or want to have my logic checked over to make sure I'm developing the right technique to solve the problem. So I guess that's what it is.

HS-Scientist and YawningDog27: I do think g(x) would be continuous along this interval, and that the zero in [0, 1] would be a relative minimum? Thank you.

Last edited: Jun 15, 2013
14. Jun 15, 2013

Infrared

Why would the zero of g be a relative min and how is that relevant to the problem? We introduced the function g because showing that it has a zero in [0,1] is equivalent to showing that f(x)=x for some x in [0,1].

15. Jun 15, 2013

Justabeginner

Using the Intermediate Value Theorem, you can conclude that for this continuous function g, on the intervals, [a, b] and [c,d ], for any z, there exists a c such that f(c)= z. If the function f is defined on the interval [0, 1] and g(x)= f(x) - x, then g can be valid on the interval [0, 1] and [-1, 1].

g(1) = f(1)- 1 and g(0) = f(0). Then g(0)= g(1) This means that there is a zero between g(0) and g(1) and on the interval [0, 1] g(c)= z= 0. And since g(x) = f(x)- x, f(x) - x= 0 and f(x)= x.

Is this logic correct? Thank you so much.

16. Jun 15, 2013

Infrared

I don't think this is what you meant to say. What are these intervals [a,b] and [c,d]? And ANY z?
What do you mean by valid? Do you mean that g can only take on values in [-1,1] (as in the sense of my above post)?
ok
Why?

Even assuming that g(1) and g(0) were equal (they're not in general), how does this follow?
Yes, that is why a zero of g corresponds to a point where f(x)=x.

Last edited: Jun 15, 2013
17. Jun 15, 2013

Justabeginner

The two intervals [a,b] and [c,d] are the intervals that are prescribed by the Intermediate Value Theorem? :S

Yes, I mean they can take on the values within that interval.

Because f(x)= x?

And I thought that this inference could be supplied by the same reason that f(x)= x.

18. Jun 15, 2013

Infrared

Why does f(x)=x?

Also, state the intermediate value theorem clearly and how you are using it.

19. Jun 15, 2013

Justabeginner

f(x)= x, because g(x)= f(x) - x, and g(x)= 0.

I think that the Intermediate Value Theorem states that for the function, f(x), which is continuous on the interval [a,b], and there exists a value Z between f(a) and f(b)- there must be at least one value Z within [a, b] such that f(c)= Z? In this problem, I'm using the Intermediate Value Theorem to prove that such a value exists between f(a) and f(b).

20. Jun 15, 2013

Infrared

Why is g(x)=0?