1. Find a solution with one corner for the problem of minima of the functional 4 v(y (x)) = { (y'− 1)² (y′+1)²dx, y (0) = 0, y (4) = 2. 1. y = -x for 0<x<1; y = x-2 for 1<x<4; and y = x for 0<x<3; y = -x+6 for 3<x<4. Along both polygonal curves the functional has a minimum.