Let f: R R be a function defined by f= e^ |x| -e^ -x e^x+e^ -x then

Step-by-step solution

The function is defined as f(x) = e^ |x| - e^ -x e^x + e^ -x . Case 1: x 0 . Then |x| = x , so f(x) = e^x - e^ -x e^x + e^ -x = (x) . For x 0 , (x) ranges from 0 to 1 (exclusive of 1 as x ). Case 2: x Thus, the function is f(x) = 0 for all x 0 and f(x) = (x) for x > 0 . Since f(x) = 0 for all x (- , 0] , the function is not injective because f(0) = f(-1) = 0 . The range of the function is [0, 1) . Since the codomain is R , the range is not equal to the codomain, so the function is not surjective. Answer: f is neith