Let the function f: R [ 1 3 , 1 ] be defined by f(x) = e^ 2x - e^x + 1 e^ 2x + e^x + 1 . Then f(x) is

Step-by-step solution

f(x) = e^ 2x - e^x + 1 e^ 2x + e^x + 1 Replacing x by -x , we get: f(-x) = e^ -2x - e^ -x + 1 e^ -2x + e^ -x + 1 = 1 - e^x + e^ 2x 1 + e^x + e^ 2x = f(x) Since f(-x) = f(x) for all x R , the function is even and hence many-one. To find the range, let t = e^x . Since x R , t > 0 . Let y = t^2 - t + 1 t^2 + t + 1 yt^2 + yt + y = t^2 - t + 1 (y-1)t^2 + (y+1)t + (y-1) = 0 For y = 1 , we get 2t = 0 t = 0 , which is rejected since t > 0 . Thus, 1 is not in the range. For y 1 , since t is real, the discriminant D 0 : (y+1