Proof of the mean value theorem

[O.J.]

the step i don't understand about the proof in this page http://en.wikipedia.org/wiki/Mean_value_theorem is the last one, where they assume that the x at which f'(x) is equal to the secant slope is the same x at which g'(x) = 0. can someone explain to me why this is so in a formal way?

 

[StatusX]

It's because they have defined g(x)=f(x)+rx, where r is the negative of the secant slope. So at all points, g'(x)=f'(x)+r.

 

[O.J.]

the thing i don't get is WHY he assumed that using rolle's theorem would give a correct answer. how did they know that converting it to a rolle's theorem suitable case where the right x is the same x at which the secant slope is equal to f'(c)?

 

[StatusX]

Are you asking how they came up with the idea to add rx to f(x) so as to convert the problem to a case where you can apply Rolle's theorem? Well, say you're given the problem:

(1) Prove that if f is differentiable on [a,b], there is some point c in [a,b] with f'(c)=s, where s is the slope of the secant line between (a,f(a)) and (b,f(b)).

(note I've called s what they've called -r) And assume further that you know Rolle's theorem, that:

(2) If f is differentiable on [a,b] and f(a)=f(b), then there is some point c in [a,b] with f'(c)=0.

Then you'll notice that (1) has a very similar structure to (2), and in fact, (1) is a special case of (2). So you'd naturally ask if there's a way to use (2) to prove (1).

(Actually, to be fair, I would probably first look at the proof of (2) to see if I could modify it to get (1). It's usually not the case that you can just use the statement of a specific theorem to prove a more general theorem, although there are other examples. You just get lucky in this case)

Next, you know the derivative of f(x)+g(x) is f'(x)+g'(x). You also might notice that if you subtract the secant line, h(x)=sx, from f(x), then f(x)-h(x) has f(a)-h(a)=f(b)-h(b) (this can be seen geometrically, since h(x) intersects f(x) at a and b, so in fact f(a)-h(a)=f(b)-h(b)=0). Then you can apply Rolle's theorem to f(x)-h(x), and see there is some point c with f'(c)-h'(c)=0, or f'(c)=h'(c)=s. I'm not sure if this answers your question.

 

[mathwonk]

in math we always convert everything to a problem of checking something is zero. that's because zero is eaiser to compute with.

so trying to find a tangent line parallel to a given line, gets changed by tilting the given line down to horizontal so the we are trying to find a line that is also horizontal.

then being horizontal means having slope zero. we know of course that being a max or min implies slope zero, so that does it.

as to why the tilted liens are parallel iff the original oens are, draw a picture. its visually obvious.

 

[matt grime]

the thing i don't get is WHY he assumed that using rolle's theorem would give a correct answer.

He didn't assume anything. Remember that the order of presentation of results in mathematics is practically in the reverse order from which they were considered. That is, the final thing in a series of little lemmas is probably what they started off wanting to prove, so they tried to do it, decided that they needed some preliminary easier results so tried to do those, and recursively got to the simplest result.

 

[O.J.]

what I am askin is, how did they know that converting it to a rolle's theorem situation and finding the x at which g'(x) = 0 will give the same x at which f'(x)=secant slope... :(

 

[StatusX]

I can't understand what you're asking. g(x)=f(x)+rx at every point (by definition), so g'(x)=f'(x)+r at every point, in particular, at x=c, where we have 0=g'(c)=f'(c)+r, so f'(c)=-r, which is the secant slope.

 

[matt grime]

The lecturer knew to do this because this is the proof in the book. The proof in the book was probably discovered by someone *just thinking about it for long enough* for them to see that an approach like this might work. Finding the proof is about experimentation - they don't just appear like magic written on a bit of paper like some pseudo-religious scripture. To see why the person had the idea in the first place *just draw a picture*. Then decide what kind of translation of coordinates will help, then try to find a function that will do this for you and allow you to invoke a theorem you already know.

 

[HallsofIvy]

what I am askin is, how did they know that converting it to a rolle's theorem situation and finding the x at which g'(x) = 0 will give the same x at which f'(x)=secant slope... :(

The same way you come up with any idea in proofs: you try this, you try that until finally something works! "They" (the first person to give this proof) did not know in advance that that would work. Probably tried it because Rolle's theorem is special, simplified, case of the mean value theorem and, often (but not always) reducing to simpler cases is a good way to try.