Question

Complete the first table so that f(x) is a function.

Complete the second table so that g(x) is not a function.

Answer 1

Answer:

$\left\begin{array}{cc}x&f(x)\\-1&7\\1&-8\\6&-9\end{array}\right$

$\left\begin{array}{cc}x&g(x)\\-1&7\\6&-8\\6&-9\end{array}\right$

Step-by-step explanation:

The given table for $f(x)$ is

$\left\begin{array}{cc}x&f(x)\\-1&--\\--&-8\\6&--\end{array}\right$

For $f(x)$ to be a function, we must complete the table in such a way  that its graph will pass the vertical line test. In order words none of the inputs should repeat.

So we can fill in the blank spaces under x, with any number apart from $-1$ and $6$.

As for the output column, any number at all can go there and $f(x)$ is still a function.

One of the possible solution is;

$\left\begin{array}{cc}x&f(x)\\-1&7\\1&-8\\6&-9\end{array}\right$

The given table for $g(x)$ is

$\left\begin{array}{cc}x&g(x)\\-1&--\\--&-8\\6&--\end{array}\right$

For $g(x)$ not to be a function, we must complete the table in such a way  that its graph will fail the vertical line test. In order words one of the inputs should repeat.

So we can fill in the blank spaces under x, with either $-1$ or $6$.

As for the output column, any number at all can go there and $g(x)$ will still not be a function.

One of the solution is;

$\left\begin{array}{cc}x&g(x)\\-1&7\\6&-8\\6&-9\end{array}\right$

Answer 2

A function assigns single output to each input.  You can use this definition to form the table.

The resultant tables would look like
$\begin{array}{cc}x&f(x)\\-1 & -2\\-4 & -8\\6 & 12\end{array}$

and

$\begin{array}{cc}x&g(x)\\-1 & -2\\-1 & -8\\6 & 12\end{array}$

What is a function?

There are two set of values. A function connects first set's elements to second set such that each element of first set is connected to only one element of second set.

How to complete the table so as to make f(x) a function and g(x) not a function?

For f(x) to be function, we assign values such that no input get two values.

$\begin{array}{cc}x&f(x)\\-1 & -2\\-4 & -8\\6 & 12\end{array}$

For g(x) to not being a function, we can assign two outputs to single input:

$\begin{array}{cc}x&g(x)\\-1 & -2\\-1 & -8\\6 & 12\end{array}$ (here -1 is assigned -2 and -8 both)

Thus, we have

The resultant tables would look like
$\begin{array}{cc}x&f(x)\\-1 & -2\\-4 & -8\\6 & 12\end{array}$

and

$\begin{array}{cc}x&g(x)\\-1 & -2\\-1 & -8\\6 & 12\end{array}$

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