# Writing a non symmetrical function as an even or odd function

I have been given the task to rewrite a function as a sum of an even and an odd function. But when I went to analyze the parts of the original function I noticed that one part was neither even nor odd.

The original function was: $$g(x) = (x + 1) / (x^2 - 3x + 4$$)

After analyzing I realized that when using f(-x) to determine symmetry it produced a 'neither' result. How would I go about writing a function that is neither even nor odd as an even function. (It would have to be an even function as the (x+1) part of the original function is odd and since I need to write it as a sum of even and odd, the denominator would need to be even.

• Perhaps you can show us a little more of your workings? It is unclear what you are stuck with.
– Matt
Oct 6, 2018 at 16:32
• Oct 6, 2018 at 16:35
• I realize that x^2 - 3x + 4 is neither even nor odd. But I must write it as an even function, how do I go about that? Do I just change the signs to make it even?
– user594350
Oct 6, 2018 at 16:35
• Hint: Complete the square with the quadratic Oct 6, 2018 at 16:36

You don’t need to analyze components of a function $$g(x)$$ individually to write it as sum of an even and an odd function.

Any function $$g(x)$$ (of course assuming the domain is symmetric) can be written as: $$g(x)=\underbrace{\frac{g(x)+g(-x)}{2}}_{\text{even}}+\underbrace{\frac{g(x)-g(-x)}{2}}_{\text{odd}}$$ The first function is even and the second is odd.

• By first function are you referring to (x+1)?
– user594350
Oct 6, 2018 at 17:19
• @EmmaPascoe I am considering the entire function $g(x)=\frac{x+1}{x^2-3x+4}$. So the even part (which is the first term in my expression written above) will be $\frac{1}{2}\left[\frac{x+1}{x^2-3x+4}+\frac{-x+1}{x^2+3x+4}\right]$. Likewise the odd part, will be $\frac{1}{2}\left[\frac{x+1}{x^2-3x+4}-\frac{-x+1}{x^2+3x+4}\right]$ Oct 6, 2018 at 17:58

Suppose we have a general function $$f(x)$$ and we want to express it as the sum of an even function and an odd function so that $$f(x)=E(x)+O(x)$$How might we find $$E(x)$$ and $$O(x)$$? Well one obvious thing would be to compute $$f(-x)=E(-x)+O(-x)=E(x)-O(x)$$ (using the definitions of even and odd)

Simply adding the two equations gives $$2E(x)=f(x)+f(-x): E(x)=\frac{f(x)+f(-x)}2$$

and subtracting gives $$2O(x)=f(x)-f(-x): O(x)=\frac{f(x)-f(-x)}2$$

So if $$E(x)$$ and $$O(x)$$ exist, they must have this form, and it is easy to check that they are even and odd functions respectively.

So the decomposition into even and odd functions, where it exists (depends where \$f(x) is defined), is unique, and can be determined by these equations.