I've been having trouble solving this:

Prove that if the function f(X) E D[0,1] and f'(0)=f'(1)=0, f(0)=1, f(1)=0, there is a X₀ E (0,1) so that f''(X₀) >= 4.

Please help.

Thanks.

What do you man by D[0,1]?

Oh, sorry. I thought it was an international symbol.

Anyway, it means differentiable in a given interval or something like that. I'm not sure that that's the word, again, my English is not perfect, but you got the idea, right?

Oh and I forgot. Differentiable twice, not once! Differentiable twice in the interval [0,1].

Thanks.

Vague thoughts: Is the mean value theory or the intermediate value theorem relevant here?

(I'd have to look up their exact form.)

Ah differentiable twice, so that means f, f' are both continuous so you can apply the mean value theorem (even twice):

f differentiable on [a,b] then there is a x E (a,b) such that
f'(x) = ( f(b)-f(a) ) / (b-a) (the mean descent/ascent).

f'(0)=f'(1)=0, f(0)=1, f(1)=0, there is a X0 E (0,1) so that f''(X0) >= 4

Applying the theorem once gives an x1 such that f'(x1)=(0-1)/(1-0)=-1.
Applying the theorem again for f' on the interval [0,x1] gives an x2 such that f''(x2)=(-1-0)/(x1-0)= -1/x1.
Applying the theorem again for f' on the interval [x1,1] gives an x3 such that f''(x3)=(0+1)/(1-x1)=1/(1-x1).

now I am stuck I think...




Greetings,

Han.

James R:

Yes, these theorems are relevant. In fact this is one of 12 problems given to me as a homework on the subject of the mean and intermediate value theorems.

.
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I was told the answer once but I forgot it :o...

I remember that it has something to do with the fact that the original function is

f(x)=2X² + C

but that's just one way to do it, I'm sure there are a dozen more.

Han, thanks for trying :)