From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Help with a math problem??? Date: 8 Aug 2000 08:10:56 -0400 Newsgroups: sci.math Summary: [missing] In article <8moqp3$pnl$1@nnrp1.deja.com>, wrote: : I got this math problem at an analytical math :course and I can't solve it? So is there anyone :who would know the answer or know where I could :search for it? It is probably easy for you but a :really hard one for me: : : f:[a,b]-> R is derivetive at [a,b] and : f'(a) < y(0) < f'(b). : Show that there is x(0) that belongs to ]a,b[ : so that f'(x(0))= y(0) : This is a standard theorem (attributed to Darboux): Suppose f is differentiable on [a, b] - with the stipulation that the derivative at b is taken from the left and the derivative at a is taken from the right (so no extension of f beyond [a, b] is needed). Suppose, as you did, that f'(a) < y(0) < f'(b), and look for x(0) such that f'(x(0)) = y(0). Motivation: We recall Fermat's theorem - that a function f which has an extreme value at an interior point c of [a, b] and which also has a derivative at c must have f'(c)=0. So, we can try reduce the given problem of prescribed derivative y(0) to the problem of prescribed derivative 0 od a related function. Now the technicalities - you finish some of the steps on your own, I trust: We invent a function g(x) = f(x) - x * y(0) observe that g'(x) = f'(x) - y(0) hence g'(a) < 0 and g'(b) > 0 . The function g is continuous on [a, b], so by the extreme value property it must have a minimum at some x(0) in [a, b]. Now pick it up: Can x(0) = a or x(0) = b? (Can f(a) or f(b) be minimal?) You can find out that x(0) must be strictly between a and b, and the conclusion follows from Fermat's Theorem. Good luck, ZVK(Slavek). ============================================================================== From: spamless@Nil.nil Subject: Re: Help with a math problem??? Date: 8 Aug 2000 08:50:18 -0400 Newsgroups: sci.math mathpath1@my-deja.com wrote: > I got this math problem at an analytical math > course and I can't solve it? So is there anyone > who would know the answer or know where I could > search for it? It is probably easy for you but a > really hard one for me: > f:[a,b]-> R is derivetive at [a,b] and ^^^^^^^^^^ should be "differentiable" > f'(a) < y(0) < f'(b). > Show that there is x(0) that belongs to ]a,b[ > so that f'(x(0))= y(0) > Sorry for my bad English skills and that I > couldn't find all the right symbols from my > computer but I think you get the idea! It looks fine - don't worry about it. Also, don't use special symbols in USENET, for it is an ascii forum. You did a good job. Of course, if f' is continuous, and f'(a)(f(b)-f(a))/(b-a) ------------- If y_0<(f(b)-f(a))/(b-a). We know f'(a)(f(b)-f(a))/(b-a) then consider g(u)=(f(b)-f(u))/(b-u)....