First off I want to apologize for bombarding this subforum with my gazillion questions. If my continuous barrage of questions poses a problem just let me know and I'll stop. 1. The problem statement, all variables and given/known data For each value of ε, find a positive value of δ such that the graph of the function leaves the window (a − δ) < x < (a + δ), (b − ε) < y < (b + ε) by the sides and not through the top or bottom. g(x) = −x^3 + 2 a = 0 b = 2 ε = 0.1, 0.01, 0.001 For ε = 0.1, δ must be less than or equal to what value? 2. Relevant equations 0<abs(x-a)<δ then abs(f(x)-b)<ε 3. The attempt at a solution abs(−x^3 + 2-2)<ε abs(−x^3)<.1 1.9<−x^3<2.1 0<abs(x-0)<δ -δ<x<δ 1.9<−x^3 -(1.9^1/3)>x −x^3<2.1 x>-(2.1^1/3) (-2.1^1/3)<=δ<x<δ<=-(1.9^1/3) δ<=-(1.9^1/3) || δ<=-1.2386 Is this right? I tried to get everything to match of properly, but I'm not sure if I did it correctly. I'm not exactly sure what I even did just now :/ Thanks in advance and sorry for all the questions.