y = 2 + 2sec(2x) The upper part of the range will be when the secant has the smallest positive value up to infinity. The smallest positive value of the secant is 1 So the minimum of the upper part of the range of y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4 So the upper part of the range is [4, ) The lower part of the range will be from negative infinity up to when the secant has the largest negative value. The largest negative value of the secant is -1 So the maximum of the lower part of the range of y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0 So the lower part of the range is (, 0]. Therefore the range is (, 0] U [4, )