# 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 show that O(x)=f(x)-E(x) is odd.

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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).**