(l1) Using only your knowlcdgc of functions and their graphs (and NOT using any calculus tcchniqucs], justify};r why (0.") is a local minimum and that cach point on the unit circle 3:2 + y2 = 1 is a local maximum. Hint: (i) Evaluolt': the function at the critical points. (ii) Consider the sign of f(1:,y). (iii) Consider the limiting behavior of f(;r:, y) as both :1: } ice and y } ice