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?

- Thread starter Syeve
- Start date

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

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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.

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

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

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

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:

- Nov 2010

- 3,728

- 1,571

Keeping the same notation as you gave originally, I would say it is the second line of my post above.

- Apr 2005

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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.