Operations with Functions

Sep 2018
4
0
Pennsylvania
Hi, I need help in using operations with functions. I'm familiar with functions of equations, (ex. f(x) = x[SUP]2[/SUP], g(x) = x + 1, f + g = x[SUP]2[/SUP]+ 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?
 
Nov 2010
3,728
1,571
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:
Apr 2005
20,249
7,909
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.