Answer to Question #161344 in Calculus for Paul

Question

consider the function f(x)=xcos(x). prove that there exists 0<x<x/2 such that f(x)=x*2/8

Answer 1

Solution. Let f(x) = X * 2/8 . Then solutions to f(x) = x cos (x) will also be solutions to f(x) = 0 < x < x/2. Notice that f(0) < x. So if we can find a number c so that f(x) ≤ 0, then we’d have that f(x) ≤ 0 ≤ f(0), and so the Intermediate Value Theorem will tell us that f(x) = 0 has a solution in between 0 and c. In fact, x = 1 will work, since (1)3 = 1, and cos(1) ≤ 1. So the solution to cos(x) = x 3 is in the interval [0, 1].