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 cant understand why do i need to know that its differentiable continuously it seem like the fact that it differentiable twice is enough ??