Question

Prove the mean value theorem using rolle's thm.

Answer 1

http://planetmath.org/encyclopedia/ProofOfMeanValueTheorem.html

Answer 2

1=2

Answer 3

very close paul

Answer 4

ho ho ho

Answer 5

assume = is an integer variable?

Answer 6

ok thats cool alchemista
thanks

Answer 7

might have a look later

Answer 8

mvt given blah blah blah there exists a number
\[c\in (a,b)\] such that
\[f'(x)=\frac{f(b)-f(a)}{b-a}\]

Answer 9

Only if f is continuous on [a,b] and differentiable on (a, b)

Answer 10

yeah that was the blah blah blah part

Answer 11

MVT relies on Rolle's, Rolle's relies on extreme value theorem.

Answer 12

gimmick is this: apply rolle's theorem to

\[h(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)\]\]

Answer 13

If you think about it you just transform the interval the MVT is being applied on so its flat

Answer 14

Then the rolle's theorem applies

Answer 15

I can show a diagram, ill find one, one second.

Answer 16

a strange looking creature that really just gives the vertical distance between the function f and the secant line

Answer 17

what alchemista said. this
\[h(x)\] just makes it flat essentially.

Answer 18

you can construct the function yourself. find the equation of the secant line using the point- slope formula and then subtract off the function. that is what you will get

Answer 19

Rolle's thm:

Let f be continuous on [a,b] and differentiable on (a,b).
Then there is c between (a,b) such that f'(c)=0.

so we have \[f'(c)=\lim_{x \rightarrow c}\frac{f(x)-f(c)}{x-c}=0\]

we want to show \[f'(c)=\frac{f(b)-f(a)}{b-a}\]

Answer 20

Yes it simple to construct. Anyways I can't find a nice diagram, sorry.

Answer 21

then check that
\[h(x)\] satisfies the hypothesis or rolle's theorem, that is check that
\[h(a)=h(b)=0\] and then go to the conclusion. say that there is some
\[c\in (a,b)\] such that
\[h'(c)=0\] which will tell you that for that particular c,
\[f'(c)-\frac{f(b)-f(a)}{b-a}=0\] and you are done

Answer 22

what is more interesting that this gimmicky proof is to use an actual example, say a nice easy cubic polynomial on some interval, and see exactly how it works in the specific case rather than the general case

Answer 23

no i only like general

Answer 24

lol

Answer 25

i dont care about examples

Answer 26

i just didn't realize all you needed to prove the mean value thm was the rolle's thm
i have seen plenty of examples

Answer 27

well that is a fine preference and of course necessary if you want a proof. but for the mechanics of why it works it is instructive to use a specific example, and actual f and an actual f', a real live a and a real live b, and an actual line and an actual derivative. it will show you much more than the abstract nonsense from the proof

Answer 28

yes but you are being cheated the whole way

Answer 29

look closely at any beginning calc text.

Answer 30

no! i said i have seen plenty of examples

Answer 31

maybe it is not clear what i am saying. but let me finish anyway about being cheated.

Answer 32

i know how it works

Answer 33

look in the book whatever calc book you have. what is the proof of mean value? depends on rolle

Answer 34

what is the proof of rolle? depends on "intermediate value theorem"

Answer 35

what is proof of intermediate value theorem?

here comes the smoke. "it seems reasonable..." or " a picture suggests..." unless you are using some real calc book like spivak, there is no actual proof. not that it is hard, but you need to use "completeness" of real line or least upper (lower) bounds to prove it.

Answer 36

so it is all built on a house of cards

Answer 37

just thought i would mention it

Answer 38

Well I don't consider a calculus text book a calculus text book unless its as good as the Spviak book

Answer 39

Spivak indeed uses the least upper bound axiom for IVT, EVT

Answer 40

Also as I said before rolles is based on EVT not IVT.