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True or False- Functions and Continuity

  1. Jun 14, 2013 #1
    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. jcsd
  3. Jun 14, 2013 #2
    Would this satisfy: f(x) = x+1 ?
     
  4. Jun 14, 2013 #3
    No it wouldn't. This is a one-to-one function.
     
  5. Jun 14, 2013 #4

    Mark44

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    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
  6. Jun 14, 2013 #5

    Office_Shredder

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    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?
     
  7. Jun 14, 2013 #6
    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.
     
  8. Jun 14, 2013 #7

    Office_Shredder

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    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
     
  9. Jun 14, 2013 #8

    LCKurtz

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    @justabeginner I am curious to know whether these questions you are posting are from a class you are taking or from self studying.
     
  10. Jun 14, 2013 #9

    Mark44

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    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.
     
  11. Jun 14, 2013 #10
    Hint: Consider the function g defined by [itex] g(x)=f(x)-x [/itex]. Note that by the constraint [itex] 0 \leq f(x) \leq 1 [/itex], [itex] 0 \leq g(0) \leq 1 [/itex] and that [itex] -1 \leq g(1) \leq 0 [/itex]. How can you use this along with g's continuity to show that [itex] g [/itex] has a zero in [itex] [0,1] [/itex]?
     
  12. Jun 14, 2013 #11
    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
  13. Jun 14, 2013 #12

    LCKurtz

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    As is usual on PF, the hints escalate until soon the problem will be completely worked.
     
  14. Jun 15, 2013 #13
    (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
  15. Jun 15, 2013 #14
    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].
     
  16. Jun 15, 2013 #15
    So I was thinking about this problem and I think I might've got something.

    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.
     
  17. Jun 15, 2013 #16
    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
  18. Jun 15, 2013 #17
    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.
     
  19. Jun 15, 2013 #18
    Why does f(x)=x?

    Also, state the intermediate value theorem clearly and how you are using it.
     
  20. Jun 15, 2013 #19
    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).
     
  21. Jun 15, 2013 #20
    Why is g(x)=0?
     
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