Compare x to the known numbers 4, 9, 11, 15, and 17. What happens to the 3rd and 4th numbers if x is less than or equal to 11? What if x is between 11 and 15? Or greater than or equal to 15?

Once you decide which two numbers will be 3rd and 4th for the range of x you are considering, write their average, set it equal to 13, and solve for x. Then check that your solution actually fits the assumed position of x in the ordered list.

In a new expression line, type m(x) = median(4, 9, 11, 15, 17, x). This tells Desmos to compute the median of the six numbers as x varies.

In another line, type y = m(x) and in a third line type y = 13. You will see the graph of the median as a function of x and a horizontal line at 13.

Use the intersection tool (click on the point where the two graphs meet) to see the x-coordinates where m(x) = 13. Among these intersection points, identify the smallest x-value; that is the least possible value of x that makes the median 13.

There are six numbers: 4, 9, 11, 15, 17, and x.

When six numbers are arranged in ascending order, the median is the average of the 3rd and 4th numbers.

Call the 3rd number a and the 4th number b. The problem says the median is 13, so

So we need the two middle numbers, a and b, to add up to 26.

First list the known numbers in order: 4, 9, 11, 15, 17.

We want to find the least possible value of x that still allows the median to be 13. That means we should try to make x as small as we can, then see if the median can reach 13.

Check what happens if x is 11 or less:

• If $x \le 9$, the ordered list starts with x, 4, 9, 11, ... so the 3rd and 4th numbers are 9 and 11.
  • Their average is $(9 + 11)/2 = 10$, which is less than 13.
• If $9 < x \le 11$, the order is 4, 9, x, 11, 15, 17, so the 3rd and 4th numbers are x and 11.
  • Then the median is $(x + 11)/2$. Even at the largest possible value in this range, $x = 11$, the median is $(11 + 11)/2 = 11$, still less than 13.

So if $x \le 11$, the median cannot be 13. Therefore, $x$ must be greater than 11.

Now consider $x > 11$ and look at how the numbers arrange.

1. If $11 < x < 15$, the order is 4, 9, 11, x, 15, 17.
   • The 3rd and 4th numbers are 11 and x.
   • The median is

Set this equal to 13:

But this solution is not in the range $11 < x < 15$ (it is exactly 15), so there is no solution with $11 < x < 15$.

2. If $x = 15$, the numbers in order are 4, 9, 11, 15, 15, 17.
   • The 3rd and 4th numbers are 11 and 15.
   • Their average is

so the median is 13.

3. For any $x > 15$, the ordered list starts 4, 9, 11, 15, ... so the 3rd and 4th numbers remain 11 and 15, and the median stays 13.

Because every $x \ge 15$ works and we are asked for the least possible value of x, the correct answer is $15$.