图中的函数 $f$ 在 $0 \le x \le 6$ 上定义如下： $f : x \mapsto \tfrac{1}{2}x^{2}\quad\text{for }0 \le x \le 2,$ $f : x \mapsto \tfrac{1}{2}x + 1\quad\text{for }2 < x \le 6.$

1. Determine the minimum and maximum values of $f(x)$ over the given domain

For $0 \le x \le 2$, the minimum is $f(0) = 0$ and maximum is $f(2) = 2$.
For $2 < x \le 6$, the function increases from $f(2) = 2$ up to $f(6) = \frac{1}{2}(6) + 1 = 4$.
Since the function is strictly increasing and continuous, the range spans from the minimum at $x=0$ to the maximum at $x=6$.

1. Identify the key points on the inverse graph by swapping the $x$ and $y$ coordinates of the endpoints of $f(x)$.

The key endpoints of $f(x)$ are $(0, 0)$, $(2, 2)$, and $(6, 4)$.
Swapping the coordinates gives the endpoints of $f^{-1}(x)$ as $(0, 0)$, $(2, 2)$, and $(4, 6)$.

2. Sketch the reflected segments, ensuring correct shapes (curve vs line) by reflecting across $y=x$.

From $x=0$ to $x=2$, the original graph is concave up (a parabola), so its reflection across $y=x$ is a concave down curve from $(0,0)$ to $(2,2)$.
From $x=2$ to $x=4$, draw a straight line from $(2,2)$ to $(4,6)$ reflecting the original linear section.

1. Find the inverse expression for the quadratic part of the function.

Let $y = \frac{1}{2}x^2$.
Then $x^2 = 2y$.
Since $x$ is in the domain $0 \le x \le 2$, $x$ is non-negative, so we take the positive square root:
$x = \sqrt{2y}$
Therefore, the first expression for $f^{-1}(x)$ is $\sqrt{2x}$.

2. Determine the valid set of $x$ values (domain) for this first expression.

The domain of $f^{-1}$ for this expression corresponds to the range of $f$ for $0 \le x \le 2$.
When $x=0, y = \frac{1}{2}(0)^2 = 0$.
When $x=2, y = \frac{1}{2}(2)^2 = 2$.
The range is $0 \le y \le 2$.
Thus, the expression $f^{-1}(x) = \sqrt{2x}$ is valid for $0 \le x \le 2$.

3. Find the inverse expression for the linear part of the function.

Let $y = \frac{1}{2}x + 1$.
$\frac{1}{2}x = y - 1$
$x = 2y - 2$
Therefore, the second expression for $f^{-1}(x)$ is $2x - 2$.

4. Determine the valid set of $x$ values (domain) for this second expression.

The domain of $f^{-1}$ for this expression corresponds to the range of $f$ for $2 < x \le 6$.
When $x \to 2, y \to \frac{1}{2}(2) + 1 = 2$.
When $x=6, y = \frac{1}{2}(6) + 1 = 4$.
The range is $2 < y \le 4$.
Thus, the expression $f^{-1}(x) = 2x - 2$ is valid for $2 < x \le 4$.