We need to find the minimum value of the expression $\max \{p^2 - 2p + 4, -p^2 + 2p - 4\}$ in the range $0 \le p \le 1$.

Analyzing the Quadratic Expressions

Let the two expressions be:

• $f_1(p) = p^2 - 2p + 4$
• $f_2(p) = -p^2 + 2p - 4$

We can rewrite $f_1(p)$ by completing the square:

$f_1(p) = (p^2 - 2p + 1) + 3 = (p-1)^2 + 3$.

This is an upward-opening parabola with vertex at $(1, 3)$.

We can rewrite $f_2(p)$ as:

$f_2(p) = -(p^2 - 2p + 4) = -( (p-1)^2 + 3 ) = -(p-1)^2 - 3$.

This is a downward-opening parabola with vertex at $(1, -3)$.

Evaluating Expressions in the Range $0 \le p \le 1$

Consider the interval $0 \le p \le 1$.

• For $f_1(p) = (p-1)^2 + 3$:
  • At $p=0$, $f_1(0) = (0-1)^2 + 3 = 1 + 3 = 4$.
  • At $p=1$, $f_1(1) = (1-1)^2 + 3 = 0 + 3 = 3$.
  • Since the vertex is at $p=1$, the function $f_1(p)$ is decreasing over the interval $[0, 1]$. The values range from 3 to 4, i.e., $f_1(p) \in [3, 4]$.
• For $f_2(p) = -(p-1)^2 - 3$:
  • At $p=0$, $f_2(0) = -(0-1)^2 - 3 = -1 - 3 = -4$.
  • At $p=1$, $f_2(1) = -(1-1)^2 - 3 = 0 - 3 = -3$.
  • Since the vertex is at $p=1$, the function $f_2(p)$ is increasing over the interval $[0, 1]$. The values range from -4 to -3, i.e., $f_2(p) \in [-4, -3]$.

In the range $0 \le p \le 1$, we observe that $f_1(p) \ge 3$ and $f_2(p) \le -3$. Therefore, $f_1(p)$ is always greater than $f_2(p)$ in this interval.

Determining the Minimum Value

The expression simplifies to:

$\max \{p^2 - 2p + 4, -p^2 + 2p - 4\} = p^2 - 2p + 4$ for $0 \le p \le 1$.

We need to find the minimum value of $f_1(p) = p^2 - 2p + 4$ in the range $0 \le p \le 1$.

As established, $f_1(p)$ is decreasing on $[0, 1]$. The minimum value occurs at the right endpoint, $p=1$.

Minimum value = $f_1(1) = (1)^2 - 2(1) + 4 = 1 - 2 + 4 = 3$.

Final Answer

The minimum value of the expression in the given range is 3.