Show that f is the sum of an even function and an odd function. f(x)=E(x)+O(x)Let E(x)=(f(x)+f(-x))/2. Show that E(-x)=E(x), so tat E is Even. Then...

For a function to be even, the following constraint must
be verified:

E(-x) =
E(x)

We'll calculate E(-x) and we'll check if replacing x
by -x, we'll get the function E(x).

E(-x) = [f(-x) +
f(-(-x))]/2

E(-x) = [f(-x) +
f(x)]/2

Since the addition from numerator is commutative,
then E(-x) = E(x), therefore the function E(x) is even.

Let
the odd function be O(x) = f(x) - E(x)

An odd function must
respect the constraint O(-x) = -O(x).

We'll replace x by -x
within the expression of O(x).

O(-x) = f(-x) -
E(-x)

But E(-x) = E(x) => O(-x) = f(-x) -
E(x)

We'll replace E(x):

O(-x)
= f(-x) - f(x)/2 - f(-x)/2

O(-x) = [2f(-x) - f(x)]/2 -
f(x)/2

O(-x) = f(-x)/2 - f(x)/2
(1)

-O(x) = E(x) - f(x)

-O(x)
= (f(x)+f(-x))/2 - f(x)

-O(x) = f(-x)/2 - f(x)/2
(2)

We notice that the expressions (1) and (2) are equal,
therefore O(-x) = O(x), so the function O(x) is
odd.

Therefore, the function f(x) is the sum
of an even and an odd functions: f(x) = E(x) +
O(x).