1. The problem statement, all variables and given/known data Find min/max/saddle point of the function : F(x,y) = y^2 - 2*y*cos(x) ; 1 <= x <= 7 F_x = 2y*sin(x) F_xx = 2y*cos(x) F_xy = 2cos(x) F_y = 2y - 2cos(x) F_yy = 2 Finding its critical points : F_x = 2y*sin(x) = 0 F_y = 2y - 2cos(x) = 0 Solving for y in F_y 2y = -2cos(x) y = -cos(x) inputing this value into F_x 2( -cos(x) ) * (sin(x) ) = 0; realizing trig identities, somehow. 2( - cos(x) ) * (sin(x) ) = -2sin(x)cos(x) = -2sin(2x) so , -2sin(2x) = 0 That function equals 0 when x = n*pi but we have boundaries so , 1<= 2n*pi <= 7 or 1/2 <= n*pi <= 7/2 the only solutions is n = 1, so n*pi = pi. ( but the book has critical points as 0,pi,2pi ) I wont do on because I know I did something wrong. I don't think they accounted for the extra 2 in 2n*pi, otherwise the critical points would be 0,pi,2pi.