Final answer:
To prove each assertion: (a) The sum of two even functions is even. (b) The sum of two odd functions is odd. (c) The sum of an even function and an odd function is neither even nor odd.
Step-by-step explanation:
To prove each assertion:
(a) Let f(x) and g(x) be two even functions. This means that f(-x) = f(x) and g(-x) = g(x) for all x in the domain. To prove that the sum of two even functions is even, we need to show that (f + g)(-x) = (f + g)(x). Using the properties of even functions, we have:
(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)
Therefore, the sum of two even functions is even.
(b) Let h(x) and k(x) be two odd functions. This means that h(-x) = -h(x) and k(-x) = -k(x) for all x in the domain. To prove that the sum of two odd functions is odd, we need to show that (h + k)(-x) = -(h + k)(x). Using the properties of odd functions, we have:
(h + k)(-x) = h(-x) + k(-x) = -h(x) - k(x) = -(h(x) + k(x)) = -(h + k)(x)
Therefore, the sum of two odd functions is odd.
(c) Let f(x) be an even function and g(x) be an odd function. To prove that the sum of an even function and an odd function is neither even nor odd, we need to show that (f + g)(-x) is not equal to (f + g)(x). Using the properties of even and odd functions, we have:
(f + g)(-x) = f(-x) + g(-x) = f(x) - g(x) = (f - g)(x)
Since (f - g)(x) has both even and odd components, it is neither even nor odd.