# Operations with Functions

• Sep 13th 2018, 05:26 PM
Syeve
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?
• Sep 13th 2018, 05:37 PM
SlipEternal
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.
• Sep 13th 2018, 06:09 PM
Syeve
Re: Operations with Functions
So in saying, f + g, how would you answer that specific question? As in, what would the answer look like?
• Sep 13th 2018, 06:11 PM
SlipEternal
Re: Operations with Functions
Quote:

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.
• Sep 13th 2018, 06:14 PM
Syeve
Re: Operations with Functions
Okay, thank you so much, it does make more sense now!
• Sep 14th 2018, 03:39 AM
HallsofIvy
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.