We are given a quadratic function $f(x) = x^2 - 2ax + 2$ defined on the interval $0 \le x \le 1$. We need to find the minimum value of this function on the given interval.

1. Problem Description

We are given a quadratic function $f(x) = x^2 - 2ax + 2$ defined on the interval $0 \le x \le 1$. We need to find the minimum value of this function on the given interval.

2. Solution Steps

The vertex of the quadratic function $f(x) = x^2 - 2ax + 2$ occurs at $x = \frac{-(-2a)}{2(1)} = a$.

We need to consider three cases:

Case 1: $a < 0$. In this case, the vertex is to the left of the interval $[0, 1]$. Since the parabola opens upwards, the minimum value will occur at the right endpoint of the interval, $x = 1$.

$f(1) = 1^2 - 2a(1) + 2 = 1 - 2a + 2 = 3 - 2a$.

Case 2: $0 \le a \le 1$. In this case, the vertex lies within the interval $[0, 1]$. Since the parabola opens upwards, the minimum value will occur at the vertex, $x = a$.

$f(a) = a^2 - 2a(a) + 2 = a^2 - 2a^2 + 2 = -a^2 + 2$.

Case 3: $a > 1$. In this case, the vertex is to the right of the interval $[0, 1]$. Since the parabola opens upwards, the minimum value will occur at the left endpoint of the interval, $x = 0$.

$f(0) = 0^2 - 2a(0) + 2 = 0 - 0 + 2 = 2$.

In summary:

If $a < 0$, then the minimum value is $3 - 2a$.

If $0 \le a \le 1$, then the minimum value is $-a^2 + 2$.

If $a > 1$, then the minimum value is $2$.

3. Final Answer

$$

\begin{cases}

3-2a, & a < 0 \\

-a^2+2, & 0 \le a \le 1 \\

2, & a > 1

\end{cases}

$$