1. ## Functions

If you have
$\displaystyle f(x)=x^{2}+3$
I know f(x) just means the function of x, or just y.

But take this:

If
$\displaystyle \frac{f(x)}{x-k}$
then
$\displaystyle r=f(k)$
Where r= remainder
This makes no since to me. some function divided by x-k yields the function of k.

How do you come up with a function?

Edit: normally f(x),f(c),f(g) are all put in the place of possible functions, but this one is not.

2. Originally Posted by integral
If
$\displaystyle \frac{f(x)}{x-k}$
then
$\displaystyle r=f(k)$
Where r= remainder
This makes no since to me. some function divided by x-k yields the function of k.
This statement is deduced based on the Polynomial Remainder Theorem. That is, if you have a function $\displaystyle f(x)$, and you divide it with another function $\displaystyle x - c$, then the remainder of the division is equals to $\displaystyle f(c)$. Proof:

Let's say you have a function $\displaystyle f(x) = (x - c)q(r) + r(x)$, where $\displaystyle q(r)$ is the quotient (the result of dividing with $\displaystyle x - c$) and $\displaystyle r(x)$ is the remainder (after dividing with $\displaystyle x - c$). According to the Polynomial Division Algorithm, the remainder of the division must have a smaller degree than the divisor, in this case, $\displaystyle x - c$. Since $\displaystyle x - c$ has a degree of 1, $\displaystyle r(x)$ must be a constant polynomial.

Now, if you substitute $\displaystyle x = c$ into the function, you'll end up with:

$\displaystyle f(c) = (c - c)q(r) + r(c)$
$\displaystyle f(c) = (0)q(r) + r(c)$
$\displaystyle f(c) = r(c)$

Therefore, the remainder is equals to $\displaystyle f(c)$.