1. The problem statement, all variables and given/known data 1. Consider the function f(x) = x^3. Prove that (a) it is not uniformly continuous on R, but that (b) it is uniformly continuous on any interval of the type [-a, a] 2. Suppose that f is uniformly continuous on a region S, and g is uniformly continuous on the region f(S). Show that the composite function g(f(x)) is uniformly continuous on S. 2. Relevant equations A function is uniformly continuous on a set S if for every e>0 exists d>0 such that |x-y|<d implies |f(x)-f(y)|<e for any x,y in S 3. The attempt at a solution 1a) I'm strugginling how to prove that it is not uniformly continuous on R. I don't really know where to start...I mean intuitively its pretty obviously its not uniformly continuous since the slope increases without bound , so no delta you choose for |x-y|<d can imply |f(x)-f(y)|<e for every x,y (due to the fact that f(x)-f(y) tends to infinity as x, y go to inifnity under the constraint x = y+d) But I really have no idea how to "prove" this..The only thing i can think of to do, but I cant htink of a reason to do it (lol) is to show that If i take |x-y|<d with x>y, d>o then i have |f(x)-f(y)| is equal to or greater than (y+d)^3 - y^3 = 3y^2d + 3d^2y + d^3 and I don't really know what to do from here. 1b) By mean favlue theorem, f(x)-f(y) = f'(c) (x-y) so |f(x)-f(y)| ≤ |f'(c)| |x-y| ≤ sup on [-a,a] of f' * |x-y| = 3a^2 |x-y| since Take d = e/3a^2 Then , if |x-y| < e/3a^2 We have |f(x)-f(y)| ≤ 3a^2 |x-y| ≤ 3a^2 * e/3a^2 = e So it has been proven 2. This one I don't even know where to begin :S Help!