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Thread: Functions

  1. #1
    Member integral's Avatar
    Joined
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    Arkansas
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    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.
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  2. #2
    Newbie
    Joined
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    Quote Originally Posted by integral View Post
    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 $\displaystyle f(c)$.
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