- #1
transgalactic
- 1,395
- 0
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
??
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
??