We have been given a rule [tex]x^{2}+x+11[/tex] and we are required to prepare a table of values using this rule and values of x from 1 to 8.
For x=1, we get: [tex]1^{2}+1+11=13[/tex]
For x=2, we get: [tex]2^{2}+2+11=17[/tex]
For x=3, we get: [tex]3^{2}+3+11=23[/tex]
For x=4, we get: [tex]4^{2}+4+11=31[/tex]
For x=5, we get: [tex]5^{2}+5+11=41[/tex]
For x=6, we get: [tex]6^{2}+6+11=53[/tex]
For x=7, we get: [tex]7^{2}+7+11=67[/tex]
For x=8, we get: [tex]8^{2}+8+11=83[/tex]
We can see that all these values are prime numbers. So, we can make a conjecture that the given rule produces prime numbers.
Now we need to find a counter example, that is, a value of x that produces a non-prime number, that is, a composite number.
Let us calculate the value of the given function at x=10. We get the value of the function as:
For x=10, we get: [tex]10^{2}+10+11=121[/tex]
We know that 121 is a composite number as it is divisible by 11.
Therefore, x=10 generates a counter example.
