Neeeeed helppppp for a question,, (not hw)

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Discussion Overview

The discussion revolves around a mathematical problem involving a twice differentiable function with specific boundary conditions. Participants are exploring the implications of these conditions on the function's second derivative, particularly in proving that the absolute value of the second derivative is greater than or equal to 4 at some point in the interval [0,1].

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the problem and asks for help in proving that |f''(x)| >= 4 for some x in [0,1], given the conditions f(0)=0, f(1)=1, f'(0)=f'(1)=0.
  • Another participant requests to see the original poster's working and questions the urgency of the request if it is not homework.
  • A subsequent reply indicates that the question may be part of a test the following day, suggesting a sense of urgency.
  • A participant attempts to provide a proof by contradiction, applying the Mean Value Theorem to derive conditions on f' and f, ultimately concluding that either f''(x) >= 4 for some x in [0, 1/2] or f''(x) <= -4 for some x in [1/2, 1].
  • Another participant challenges the validity of the previous argument by presenting a counterexample of a function that satisfies f'' = 0 on [0, 1/2], which would not meet the initial condition, thus questioning the completeness of the proof provided.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the proposed proof and the conditions necessary to satisfy the problem. There is no consensus on the correctness of the arguments presented, and the discussion remains unresolved.

Contextual Notes

The discussion includes assumptions about the behavior of the second derivative and the application of the Mean Value Theorem, which may not be universally applicable without further justification. The counterexample presented raises questions about the completeness of the initial proof.

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neeeeed helppppp for a question,,urgent!(not hw)

Hi, I have been working on this question for some time,, but still couldn't solve it,,does anyone have an idea? thanks for helping!

Prove that if f is a twice differentiable function with f(0)=0 and f(1)=1 and f '(0)=f '(1)=0, then / f ' ' (x) / >=4 for some x in [0,1].
In more picturesque terms: A particle which travels a unit distance in a unit time, and starts and ends with velocity 0, has at some time an acceleration>=4

Hint:Prove that either f ' ' (x)>4 for some x in [0, 1/2] or f ' '(x)<-4 for some x in [1/2, 1].

and show that in fact we must have/ f ''(x)/ >4 for some x in [0,1]
 
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Perhaps you could show some of your working?

And how urgent can it be if it's not "hw"?
 
J77 said:
Perhaps you could show some of your working?

And how urgent can it be if it's not "hw"?

it will probably be a test question on the test tmr...

using contradition? or not..i don't have enough time to think about it now, just want to see some clues.
 
Last edited:
i m not sure if my answer is completed,, can anyone check it for me? thx!
suppose f''(x)<4 for all x in [0, 1/2]，apply mean value theorem,if0<=x<=1/2
(f'(x)-f'(0))/(x-0)<4 ==> f'(x)<4x,one more time mean value theorem,if0<=x<=1/2
(f(x)-f(0))/(x-0)<4x ==>f(x)<4x^2 f(1/2)<1
suppose f''(x)>-4 for all x in [1/2,1]
(f'(1)-f'(x))/(1-x)>-4 ==> f'(x)<4(1-x)
(f(1)-f(x))/(1-x)<4(1-x)=>f(x)>1-4(1-x)^2 f(1/2)>1 contradition
therefore, either f''(x)>=4 for some x in [0, 1/2]，orf''(x)<=-4 for some x in [1/2,1]
 
That can't possibly be right.

Consider an f satisfying f''=0 identically on [0,1/2]. Obviously it fails the first half of your condition. And it is easy to complete it to a smooth function satisfying the boundary conditions with f'' =>0.
 

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