# Neeeeed helppppp for a question,, (not hw)

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]

## Answers and Replies

Perhaps you could show some of your working?

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

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 dont 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]

matt grime
Science Advisor
Homework Helper
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.