Determining Whether a Function is Odd or Even from its Equation 
I don't understand how this "Equation" is even being solved.
I understand I have to substitute -x in for x but after that I don't know understand what's going on here. 
Please someone explain to me step by step. Thank you
 A: $$f(x)=x^3-x\Rightarrow f(-x)=(-x)^3-(-x)=-x^3-(-x)=-(x^3-x)=-f(x)$$
If $f(-x)=-f(x)$ then from definition $f(x)=x^3-x$ is odd
A: Remember that when we write something like 
$$
f(x) = x^3 - x
$$
what we mean is that no matter what input $x$ we put in${}^1$, then we compute the value of $f$ by substituting it into that formula. So for example,
$$
f(2) = 2^3 - 2 = 8 - 2 = 6
$$
and
$$
f(-492) = (-492)^3 - (-492) = -119,094,996
$$
and even
$$
f(\spadesuit) = \spadesuit^3 - \spadesuit
$$
So in particular, if we want to compute $f(-x)$, this is given by substituting $(-x)$ into the original formula wherever we saw an $x$. If this seems confusing (too many $x$'s?) Consider looking at $f(y) = y^3 - y$---the same function, really, and substitute in $y = (-x)$. This gives
$$
f(-x) = (-x)^3 - (-x) = (-1)^3x^3 + x = -x^3 + x = -(x^3 - x) = -f(x)
$$
where it should be noted that it is exceedingly important to pay attention to the brackets.
Now, a function is odd if, for every $x$, we have that the equality
$$
f(-x) = -f(x)
$$
which is exactly what we have shown. So this function is odd.
Now, it's probably good to see an example where this fails. Let's look at $g(y) = y^3 + y^2$. Let's compute $g(-y)$. This is
$$
g(-y) = (-y)^3 + (-y)^2 = (-1)^3y^3 + (-1)^2y^2 = -y^3 + y^2
$$
However, this function is neither $g(y)$ nor $-g(y)$, and so this function is neither even nor odd.



*

*something something domain something something

