# Even Odd or Neither

If f(x) and g(x) are both even functions, is f + g even? If f(x) and g(x) are both odd functions, is f + g odd? What if f(x) is even and g(x) is odd?

Now intuitively I get that for the first one, the resulting answer will be EVEN

For second one it will be ODD

And for the last one it will be Neither.

But I don't know how to justify that... it just came to me intuitively, Can someone show me a way through al-gebra to prove my claim.

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use the fact (f+g)(x)=f(x)+g(x) and the definition of odd/even function :-) – Bman72 Jan 15 '14 at 20:35
The easiest way to prove the 3rd one is to use an example. $f(x)=1$ and $g(x)=x$. You can prove that $h(x)=x+1$ is neither odd nor even by comparing with $h(-x)$. – ex0du5 Jan 15 '14 at 20:38

## 2 Answers

For the first two, just use the definition of even and odd to draw the desired conclusion. You will not be able to justify your conclusion on the last one. The sum of an even function and an odd function maybe even, odd, both, or neither. Bear in mind that the constant 0 function is both even and odd, as that should help you construct explicit examples for each of the four possibilities. For example, consider $f(x)=x^2$ and $g(x)=x^3.$ Then $f$ is even and $g$ is odd, but $$(f+g)(-1)=0\neq2=(f+g)(1)$$ and $$(f+g)(-1)=0\neq-2=-(f+g)(1),$$ so $f+g$ is neither even nor odd. I leave the other three cases to you.

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oh okay I see.. so if f(x) was the zero function the last problem will be an odd function.... but if g(x) was the zero function it would be an even? – Naqvi Jan 15 '14 at 20:43
Precisely correct! – Cameron Buie Jan 15 '14 at 20:45
Actually,we can say a bit more. Assuming that $f$ and $g$ are functions $E\to\Bbb R$ for some $E\subseteq\Bbb R,$ where $f$ is even and $g$ is odd, we have that $f+g$ is even if and only if $g$ is the constant $0$ function on $E$, and that $f+g$ is odd if and only if $f$ is the constant $0$ function on $E.$ That's a bit more complicated to prove, though. – Cameron Buie Jan 15 '14 at 20:51

Let $f(x)$ and $g(x)$ be even functions. Then we have $$(f+g)(-x) = f(-x) + g(-x) = f(x)+g(x) = (f+g)(x)$$

Let $f(x)$ and $g(x)$ be odd functions. Then we have $$(f+g)(-x) = f(-x)+g(-x) = -f(x) -g(x) = -(f+g)(x)$$

Let $f(x)$ be an even function and $g(x)$ be an odd function. Then we have $$(f+g)(-x) = f(-x)+g(-x) = f(x) - g(x) = (f-g)(x)$$ We can't say whether $f-g$ is even or odd.

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This only transforms the 3rd problem to an equivalent intuition. – ex0du5 Jan 15 '14 at 20:36
Oh okay that makes sence... as we are assuming here... okay what you did. Alright thanks ! – Naqvi Jan 15 '14 at 20:38
@ex0du5.. what do you mean? – Naqvi Jan 15 '14 at 20:38
I suspect that ex0du5 means that nothing has actually been shown by your manipulations in the third case. – Cameron Buie Jan 15 '14 at 20:40
@CameronBuie.. really? but it shows that (f-g)(x) is nether -(f+g)(x) nor is it (f+g)(x).... hence neither even nor odd ? – Naqvi Jan 15 '14 at 20:42