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Thread: Operations with Functions

  1. #1
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    Operations with Functions

    Hi, I need help in using operations with functions. I'm familiar with functions of equations, (ex. f(x) = x2, g(x) = x + 1, f + g = x2+ x + 1),
    but what I really need help in is understanding a problem like this:

    f = {(-3, 1), (0, 4), (2, 0)}
    g ={(-3, 2), (1, 2), (2, 6), (4, 0)}

    What is f + g? f * g? g / f?
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  2. #2
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    Re: Operations with Functions

    I'm guessing a little because I have not encountered this notation before.

    $f+g=\{(-3,3),(2,6)\}$

    $f\times g=\{(-3,2),(2,0)\}$

    $\dfrac{g}{f}=\{(-3,2)\}$

    You choose the x-components that are the same in both functions. Then you apply the operation to the y-component of just those pairs (provided the operation is well-defined). So the pairs share x-components -3 and 2, so only those pairs are important (the domain is restricted by both functions). Then in this case, the operation is well defined for the y-components unless there is division by zero. Basically, you are limiting the domain of the combined function to where both functions exist.

    Here is another way to write the functions (hopefully more consistent with how you have seen other functions written):

    $$f:\{-3,0,2\} \to \{0,1,4\}$$

    is defined by:

    $$f(x) = \begin{cases}1, & x=-3 \\ 4, & x=0 \\ 2, & x=2\end{cases}$$

    and

    $$g:\{-3,1,2,4\} \to \{0,2,6\}$$

    is defined by:

    $$g(x) = \begin{cases}2, & x < 2 \\ 6, & x=2 \\ 0, & x=4\end{cases}$$

    So, $f(-3)=1$, $f(0)=4$, $f(2)=0$, $g(-3)=2$, $g(1)=2$, $g(2)=6$, $g(4)=0$.

    Then,

    $(f+g)(-3) = f(-3)+g(-3) = 1+2 = 3$
    $(f+g)(0) = f(0)+g(0) = 4+\text{undefined} = \text{undefined}$
    $(f+g)(1) = f(1)+g(1) = \text{undefined} + 2 = \text{undefined}$
    $(f+g)(2) = f(2)+g(2) = 0+6 = 6$
    $(f+g)(4) = f(4)+g(4) = \text{undefined} + 0 = \text{undefined}$

    For any other values of $x$, both $f$ and $g$ are undefined, so the sum is undefined.
    Last edited by SlipEternal; Sep 13th 2018 at 05:56 PM.
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  3. #3
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    Re: Operations with Functions

    So in saying, f + g, how would you answer that specific question? As in, what would the answer look like?
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  4. #4
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    Re: Operations with Functions

    Quote Originally Posted by Syeve View Post
    So in saying, f + g, how would you answer that specific question? As in, what would the answer look like?
    Keeping the same notation as you gave originally, I would say it is the second line of my post above.
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  5. #5
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    Re: Operations with Functions

    Okay, thank you so much, it does make more sense now!
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  6. #6
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    Re: Operations with Functions

    This is really a matter of understanding the notation. "f = {(-3, 1), (0, 4), (2, 0)}" means f(-3)= 1, f(0)= 4, f(2)= 0 and f is not defined for any other values of x. Similarly "g ={(-3, 2), (1, 2), (2, 6), (4, 0)}" means g(-3)= 2, g(1)= 2, g(2)= 6, and g(4)= 0 and g is not defined for any other values of x.

    f(-3)= 1 and g(-3)= 2 so f+ g(-3)= f(-3)+ g(-3)= 1+ 2= 3, fg(-3)= f(-3)g(-3)= 1(2)= 2, and f/g(-3)= f(-3)/g(-3)= 1/2.

    f(0)= 4 but g(0) is not defined so f+ g, fg, and f/g are not defined at x= 0.

    f(2)= 0 and g(2)= 6 so f+ g(2)= f(2)+ g(2)= 0+ 6= 6, fg(2)= f(2)g(2)= 0(6)= 0, and f/g(2)= f(2)/g(2)= 0/6= 0.
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