Proving the Mean Value Theorem: Limiting θ(h) to 1/2

[kazuyak]

Suppose that the conditions for the Mean Value Theorem hold for the function
f : [a, a + h] → R, so that for some θ ∈ (0, 1) we have f (a + h) − f (a) = hf ′ (a + θh).
Fix f and a, and for each non-zero h write θ(h) for a corresponding value of θ.
Prove that if f ′′ (a) exists and is non-zero then lim(h→0) θ(h) = 1/2 .

I have no clue how to handle this problem. Could anyone please give me some hints?

 

[HallsofIvy]

The second derivative is, of course, the derivative of the first derivative. That is,
[tex]\frac{d^2f}{dx^2}(a)= \lim_{h\to 0} \frac{f'(a+h)- f'(a)}{h}[/itex]<br /> <br /> But you know that $hf'(a+\theta h)= f(a+ h)- f(a)$ so that $f'(a+ \theta h)= (f(a+h)- f(a))/h$.<br /> <br /> Replace f'(a+h) and f'(a) with variations on that.[/tex]

 

[kazuyak]

Do you mean that from f'(a+θh)=(f(a+h)-f(a))/h,
we get that f'(a)=(f(a+(1-θ)h)-f(a-θh))/h, f'(a+h)=(f(a+(2-θ)h)-f(a+(1-θ)h))/h
and than get f''(a) using l'hopital's rule? I doubt I'm doing what you mean, since it's leading me nowhere. Could you please give me some further hints?