Question

@dan815

Answer 1

F'(x) = f(x)
if \(c\in (x_{i-1}, x_i)\) , then \(f(c) = \dfrac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}\)
why??

Answer 2

I know by mean theorem, we have \(f'(c) = \dfrac{f(x_i) -f(x_{i-1})}{x_i-x_{i-1}}\) but don't understand how to converse to the expression above

Answer 3

So by MVT do we have

\[F'(c) = \dfrac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}\]

?

Answer 4

I don't see how its any different

Answer 5

oh its not there exists... I see

Answer 6

@rational yes
@zzr0ck3r they are different

Answer 7

simply replace F'(c) by f(c) and we're done i guess ?

Answer 8

they are same.

Answer 9

F'(x) = f(x)



plugin x = c

Answer 10

yes, :(

Answer 11

but the IMV is there exists c, not for all c. correct?

Answer 12

MVT is existence only, yes

Answer 13

Thank you. How stupid I am. hahaha..

Answer 14

this is just an example of how notation can be confusing ;)

Answer 15

true haha they shouldn't have used two different letters

Answer 16

you guys still in finals? I am done !