Differantiation proof question

[transgalactic]

f(x) is a differentiably continues on [-1,1]
and differentiable twice on (-1,1)
suppose f(0)=-1 and f(1)=1 and function f(x) has on (-1,0) an extreme point.

prove that there is a point c on (-1,1) so f''(c)>=1
??

prove:
because of ferma law
if a function has an extreme point on (-1,0) on the point c1 in (-1,0) then
f(x) differentiable on (c1) and f'(c)=0

because of mean value theorem there is a point c2 on (0,1):
f'(c2)=[f(1)-f(0)]/(1-0)=2
and because f'(x) differentiable once more we can do MVT again
f''(c3)=[f'(c2)-f'(c1)]/c2-c1=(2-0)/(c2-c1)
c1 and c2 are inside (-1,1) so the difference between them is smaller then 2
f''(c3)=2/(c2-c1)>=2/2 >=1

i can't understand why do i need to know that its differentiable continuously
it seem like the fact that it differentiable twice is enough
??

 

[CompuChip]

To apply the MVT to f' you need that f' is differentiable on (-1, 1) AND continuous on [-1, 1].
This is precisely equivalent to continuously differentiable on [-1, 1] and twice differentiable on (-1, 1), as you can check.

 

[transgalactic]

so we need f(x) to be continuously differentiable in order to do mvt for f'(x)

but we did it twice
we have not been given that f'(x) continuously differentiable too
so we didnt have the right to do mvt on f''(x)
??

 

[CompuChip]

Check again the requirements for the MVT.

Let f : [a, b] → R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), where a < b

We need f to be continuous on the closed interval and differentiable on the open interval. There are no requirements on the continuity of the derivative. So if you want to apply the MVT to f', it is enough if f' is continuous on the closed interval and differentiable on the open interval, which means that f must be continuously differentiable on the closed interval and twice differentiable (without requirement on the continuity of the second derivative) on the open interval.

 

[transgalactic]

thanks :)