If $$x$$ is a real number such that $$\max(\min(x, 2 - x), x - 4, 2x - 8) = \pi - 3$$, then the number of possible values of $$x$$ is

Let $$g(x) = \min(x, 2-x)$$, which peaks at 1 (when $$x = 1$$). For large $$|x|$$, $$g$$ goes to $$-\infty$$.

We need $$\max(g(x), x-4, 2x-8) = \pi - 3 \approx 0.142$$.

• $$g(x) = \pi - 3$$ gives $$x = \pi - 3$$ (for $$x \le 1$$) and $$x = 5 - \pi$$ (for $$1 < x \le 2$$). For each, check the other terms are smaller — yes.
• $$2x - 8 = \pi - 3$$ gives $$x = \dfrac{\pi + 5}{2} \approx 4.07$$. Then $$x - 4 \approx 0.07$$ and $$g(x) < 0$$. Max $$= \pi - 3$$. ✓
• $$x - 4 = \pi - 3$$ gives $$x = \pi + 1 \approx 4.14$$. Then $$2x - 8 \approx 0.28 > \pi - 3$$, fails.

Total 3 values.