Neeeeed helppppp for a question,, (not hw)

[antibody]

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]

 

[J77]

Perhaps you could show some of your working?

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

 

[antibody]

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.

 

[antibody]

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]

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.