True or False- Functions and Continuity

  • #1

Homework Statement


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.


Homework Equations





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.
 

Answers and Replies

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

No it wouldn't. This is a one-to-one function.
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?
 
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  • #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?
 
  • #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.
 
  • #7
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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.


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
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.
 
  • #9
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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.
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
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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]?
 
  • #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.
 
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  • #12
LCKurtz
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As is usual on PF, the hints escalate until soon the problem will be completely worked.
 
  • #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.
 
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  • #14
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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.
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
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.
 
  • #16
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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.
I don't think this is what you meant to say. What are these intervals [a,b] and [c,d]? And ANY 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].
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)?
g(1) = f(1)- 1 and g(0) = f(0).
ok
Then g(0)= g(1)
Why?

This means that there is a zero between g(0) and g(1) and on the interval [0, 1] g(c)= z= 0.
Even assuming that g(1) and g(0) were equal (they're not in general), how does this follow?
And since g(x) = f(x)- x, f(x) - x= 0 and f(x)= x.
Yes, that is why a zero of g corresponds to a point where f(x)=x.

Is this logic correct? Thank you so much.
 
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  • #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.
 
  • #18
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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.

Why does f(x)=x?

Also, state the intermediate value theorem clearly and how you are using it.
 
  • #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
g(0)=0, or at least that's what I (incorrectly?) presumed.
 
  • #22
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Yes, I don't see why that should be true. If it were then the problem would be trivial. g(0)=0 means that f(0)=0, solving the problem immediately with no need for the intermediate value theorem or any such technique. But even if g(0) was 0, how does g(1)=0 follow?
 
  • #23
Oh alright. I will omit that then. Thank you so much HS-Scientist.
 
  • #24
Infrared
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No problem, glad to help. Posts 10 and 11 give you nearly a complete solution. See if you can finish it from there.
 

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