An Impossible Solution

  • Thread starter JPBenowitz
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  • #1
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So, yesterday in class we were asked to try and solve the following problem:

Given a, b [itex]\in[/itex] R with a < b, draw the graph of an example of a continuous function f such that f: [a,b] [itex]\rightarrow[/itex] R, f(a) = f(b), and there does not exist c [itex]\in[/itex] (a, b) such that f'(c) = 0.

Now in class we arrived at the conclusion that there exists no such solution, however I beg to disagree.

Let's suppose that a = -1 and b = 1 which satisfies the inequality. Now for f(a) to = f(b) one function comes to mind, the associated power series:

f(x) = 1 + 2x + 4x2 + 8x3 + ... + 2nxn = [itex]\frac{1}{1-2x}[/itex]

Such that, f2(1) = 1 and f2(-1) = 1 therefore, f(a) = f(b).

= [itex]\frac{d}{dx}[/itex]([itex]\frac{1}{1-2x}[/itex])2

= ([itex]\frac{-4}{2x - 1}[/itex])3

Therefore, there does not exist f'(c) = 0

If we take the limit as x [itex]\rightarrow[/itex] [itex]\infty[/itex] then f'(x) = 0 however there is no particular element c of (a, b) s.t. f'(c) = 0.

My question is am I right? If not where did I go horribly wrong.
 
Last edited:

Answers and Replies

  • #2
192
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Going too fast ?
Differentiation that power series and see that at x=0
f ' (x) = 0
 
  • #3
144
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Going too fast ?
Differentiation that power series and see that at x=0
f ' (x) = 0
I accidently posted before I finished lol.
 
  • #4
gb7nash
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This statement is indeed true.

I'm not sure why you're taking a limit as x goes to infinity. In any case, if you think the statement is wrong, you need to show there exists a c in (a,b) such that f'(c) = 0 for every f(x). Unfortunately, showing one example doesn't prove anything.

Given a, b [itex]\in[/itex] R with a < b, draw the graph of an example of a continuous function f such that f: [a,b] [itex]\rightarrow[/itex] R, f(a) = f(b), and there does not exist c [itex]\in[/itex] (a, b) such that f'(c) = 0.
Consider the example f(x) = |x| and a = -1, b = 1. Does this satisfy the hypothesis? Is the conclusion satisfied?
 
  • #5
144
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This statement is indeed true.

I'm not sure why you're taking a limit as x goes to infinity. In any case, if you think the statement is wrong, you need to show there exists a c in (a,b) such that f'(c) = 0 for every f(x). Unfortunately, showing one example doesn't prove anything.



Consider the example f(x) = |x| and a = -1, b = 1. Does this satisfy the hypothesis? Is the conclusion satisfied?
Which statement is true? That there is no solution? Or that there is a solution?
 
  • #6
gb7nash
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This statement:

Given a, b [itex]\in[/itex] R with a < b, draw the graph of an example of a continuous function f such that f: [a,b] [itex]\rightarrow[/itex] R, f(a) = f(b), and there does not exist c [itex]\in[/itex] (a, b) such that f'(c) = 0.
It is possible to construct an example of a continuous function f such that there does not exist a c [itex]\in[/itex] (a,b) such that f'(c) = 0. Consider f(x) = |x|.
 
  • #7
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This statement:



It is possible to construct an example of a continuous function f such that there does not exist a c [itex]\in[/itex] (a,b) such that f'(c) = 0. Consider f(x) = |x|.
f(x) is discontinuous at 0.
 
  • #8
gb7nash
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f(x) is discontinuous at 0.
f(x) = |x| is continuous at x=0. However, it isn't differentiable at x=0.
 
  • #9
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f(x) = |x| is continuous at x=0. However, it isn't differentiable at x=0.
Sorry, that's what I meant.
 
  • #10
AlephZero
Science Advisor
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In your OP, f(x) = 1/(1-2x) is not continuous (in fact it is not even defined) when x = 1/2.
 

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