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

kazuyak
Messages
3
Reaction score
0
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?
 
Physics news on Phys.org
The second derivative is, of course, the derivative of the first derivative. That is,
\frac{d^2f}{dx^2}(a)= \lim_{h\to 0} \frac{f&#039;(a+h)- f&#039;(a)}{h}[/itex]<br /> <br /> But you know that hf&amp;#039;(a+\theta h)= f(a+ h)- f(a) so that f&amp;#039;(a+ \theta h)= (f(a+h)- f(a))/h.<br /> <br /> Replace f&#039;(a+h) and f&#039;(a) with variations on that.
 
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?
 

Similar threads

Replies
14
Views
2K
Replies
5
Views
3K
Replies
1
Views
1K
Replies
11
Views
2K
Replies
3
Views
1K
Replies
1
Views
2K
Back
Top