What is the range of the function y = - x2 + 1?

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,)