# Thread: Even and Odd Functions

1. ## Even and Odd Functions

I am correcting my tests for a little review for my final on Thursday and got hung up on this question I got wrong when I took it.

Here goes nothing:

Algebraically determine whether the following function is odd, even or neither.
F(x) = (sin(x)/(x^2)+5) - tan(x)

f(-x) = (sin(-x)/(-x^2)+5) - tan(-x)
f(-x) = (-sin(x)/(x^2)+5) + tan(x)

f(-x) does not equal f(x) so it cant be equal, now to check if its odd.
if -(f(x)) = f(-x), it will be odd.

-(f(x)) = -((-sin(x)/(x^2)+5) +tan(x))
-(f(x)) = (sin(x)/(-x^2)-5) - tan(x))
-(f(x)) = (sin(x)/(x^2)-5) - tan(x)

Thus -(f(x)) does NOT equal f(-x).
So its neither.

Or am I missing something.

2. Be careful with your notation. Exponentiation takes precedence over negation, so $-x^2$ is usually interpreted $-(x^2),$ not $(-x)^2.$

Originally Posted by Byud
if -(f(x)) = f(-x), it will be odd.

-(f(x)) = -((-sin(x)/(x^2)+5) +tan(x))
-(f(x)) = (sin(x)/(-x^2)-5) - tan(x))
-(f(x)) = (sin(x)/(x^2)-5) - tan(x)
I am not sure what you are doing here, as you start off by assuming that $-f(x)=f(-x)$ (which is what you are trying to show). To show that $f$ is odd, you should do

$f(x)=\frac{\sin x}{x^2+5}-\tan x$

$\Rightarrow-f(x)=-\left(\frac{\sin x}{x^2+5}-\tan x\right)$

$=-\frac{\sin x}{x^2+5}+\tan x$

$=\frac{-\sin x}{x^2+5}+\tan x$

$=\frac{\sin(-x)}{(-x)^2+5}-\tan(-x)$

$=f(-x).$