From http://learningcurves.blogspot.com -------- Original Message -------- Subject: From the Calculus Exam Date: Mon, 12 Sep 2005 17:24:33 GMT From: <Rudbeckia Hirta> Or, "How's that plan to learn nothing in my class working for you, Mr. I-Took-Calculus-in-High-School?" *Question:* Precisely state the Intermediate Value Theorem. *Answers:* In a continuous function, if there are 3 points a, b, and c where a>c, there should be a point b in between a and c so that a>b>c. Given a number a. IVT converts a to the last greatest integer a is divisible by. Between any two points on a graph both L and the limit x -> L exist and can be formed. If 2 numbers are substituted for x at the same distance away from c, then L is the intermediate number. If you have a f(x) and domain (a,b) and the lim x->a f(x) = f(a) and lim x->b f(x) = f(b). If you have one point c between (a,b) then lim x->c f(x) is between f(a) and f(b). Take the intermediate value to find where the function is discontinuous using [[x]]. The IVT states that if a function is not continuous at every number in its domain, then it is not continuous. For every function that is continuous through the given points f(a) and f(b) and has a defined limit, there exists a real number N anywhere between the points f(a) and f(b). On a function f(x) all values of x on the functions domain will be continuous. lim x->a f(x) = L In a closed interval when both f(a) and f(b) exists and are continuous, there exists a constant c between f(a) and f(b) which is also continuous. If a function is continuous on an interval [a,b] there is at least one point c where a < c < b such that there exists f(c)=N. -- Thane Plambeck http://www.plambeck.org/ehome.htm