Modulus Minima Maxima

We are not taking only one case, we are taking the case that actually gives a valid solution.

When you solve |2x − 6| = |x − 9|, there are two cases:

1. 2x − 6 = x − 9 → x = −3
2. 2x − 6 = −(x − 9) → x = 5

Now check both in the original expression max(|2x − 6|, |x − 9|):

At x = −3:

|2x − 6| = 12, |x − 9| = 12 → max = 12

At x = 5:

|2x − 6| = 4, |x − 9| = 4 → max = 4

We want the minimum value of the maximum, so we choose the value which gives the smaller maximum, which is x = 5.

So both cases are considered, but only one gives the minimum, that’s why the solution focuses on that case.