So we can start with the full of possibilities and eliminate them one by one.
The full set is {0,1,2,3,4,5,6,7,8,9}.
Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:
{1,3,5,7,9}
We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.
{1,3,7,9}
We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:
Remember that the range is the set of all y-values. Thus, since the minimum of the function is y=-1, then our range is . In interval notation, we use brackets to show what's included and parentheses to show what's not included.
learn this... if ur equation ends in a variable equaling a number, then there is one solution. if ur equation ends in something not equal, like 2 = 4, or 4 = 6, then there is 0 solutions. if ur equation ends in something equal to something,(the same) like 2 = 2, or 4x = 4x, then there is infinite solutions