函数 $\mathrm{f}$ 定义为
$\mathrm{f}(x) = \begin{cases} 4 - x & \text{for } 1 \le x < 3, \\ (x - 4)^2 & \text{for } 3 \le x < 4, \end{cases}$
且已知对于所有实数 $x$，恒有 $\mathrm{f}(x - 3) = \mathrm{f}(x)$。

1. State the condition that fails for the existence of an inverse

The equation $\mathrm{f}(x - 3) = \mathrm{f}(x)$ indicates that $\mathrm{f}$ is periodic. Therefore, multiple values of $x$ map to the same $y$-value, meaning $\mathrm{f}$ is a many-to-one function. A function must be one-to-one to have an inverse.

1. Determine the shape of the graph for the fundamental domain $1 \le x < 4$.

For $1 \le x < 3$, the function is $y = 4 - x$, which represents a straight line segment starting from a solid dot at $(1, 3)$ and ending at $(3, 1)$.
For $3 \le x < 4$, the function is $y = (x - 4)^2$. Since $(3-4)^2 = 1$, the curve connects continuously at $(3, 1)$. It is a parabolic curve opening upwards, ending at an open circle at $(4, 0)$.

2. Use the periodic property to sketch the domain $4 \le x < 6$.

The condition $\mathrm{f}(x - 3) = \mathrm{f}(x)$ indicates that the function is periodic with a period of $3$.
For $4 \le x < 6$, we translate the initial line segment piece ($1 \le x < 3$) exactly $3$ units to the right.
This translates the segment from $(1, 3) \to (3, 1)$ to become a line segment starting at a solid dot at $(4, 3)$ and ending at an open circle at $(6, 1)$.

3. Use the periodic property to sketch the domain $-1 < x < 1$.

For $-1 < x < 1$, we translate the corresponding piece of the fundamental graph $3$ units to the left (i.e. the portion from $2 < x < 4$).
For $-1 < x < 0$, shifting $y = 4 - x$ ($2 < x < 3$) left gives a line segment from an open circle at $(-1, 2)$ to $(0, 1)$. Thus, the $y$-intercept is $1$.
For $0 \le x < 1$, shifting $y = (x - 4)^2$ ($3 \le x < 4$) left gives a parabolic curve from $(0, 1)$ to an open circle at $(1, 0)$.
The overall graph features jump discontinuities at $x = 1$ and $x = 4$.

1. Use the periodic property of the function to find an equivalent argument within the base domain $[1, 4)$.

Given $\mathrm{f}(x - 3) = \mathrm{f}(x)$, the function is periodic with a period of 3. We can subtract multiples of 3 from 2017 to find a value within the defined domain $1 \le x < 4$.
$2017 \div 3 = 672 \text{ remainder } 1$
$2017 = 3(672) + 1$
Therefore:
$\mathrm{f}(2017) = \mathrm{f}(2017 - 3(672)) = \mathrm{f}(1)$

2. Evaluate the function using its piecewise definition.

For $x = 1$, the function falls into the first interval $1 \le x < 3$, where $\mathrm{f}(x) = 4 - x$.
$\mathrm{f}(1) = 4 - 1 = 3$

1. Sketch the graph of $y = \mathrm{g}(x)$ based on its piecewise definition.

For $1 \le x < 3$, $\mathrm{g}(x) = 4 - x$. This forms a line segment connecting $(1, 3)$ and $(3, 1)$.

For $3 \le x < 4$, $\mathrm{g}(x) = (x - 4)^2$. This is a parabolic curve starting from $(3, 1)$ and decreasing to an open circle at $(4, 0)$.

2. Sketch the graph of $y = \mathrm{g}^{-1}(x)$ by reflecting $y = \mathrm{g}(x)$ in the line $y = x$.

The line segment from $(1, 3)$ to $(3, 1)$ has the equation $y = 4 - x$. Because this line is symmetric about $y = x$, it reflects exactly onto itself.

The curve from $(3, 1)$ to the open circle at $(4, 0)$ reflects to a curve starting from $(1, 3)$ and going upwards to an open circle at $(0, 4)$.

3. Determine the values of $x$ where $\mathrm{g}(x) = \mathrm{g}^{-1}(x)$ from the sketch.

The equation $\mathrm{g}(x) = \mathrm{g}^{-1}(x)$ is satisfied at the points where the two graphs intersect or overlap.

From the sketch, the graphs of $y = \mathrm{g}(x)$ and $y = \mathrm{g}^{-1}(x)$ overlap perfectly along the shared line segment from $x = 1$ to $x = 3$.

Therefore, the functions are equal for $1 \le x \le 3$.

1. State the range of the inner function $\mathrm{g}^{-1}$ and the domain of the outer function $\mathrm{h}$.

The range of the inverse function $\mathrm{g}^{-1}$ is the domain of $\mathrm{g}$:
$R_{\mathrm{g}^{-1}} = D_{\mathrm{g}} = [1, 4)$
The domain of $\mathrm{h}$ is given by its piecewise definition:
$D_{\mathrm{h}} = [0, 1) \cup [1, 3] = [0, 3]$

2. Evaluate the condition for the existence of the composite function.

For the composite function $\mathrm{h}\mathrm{g}^{-1}$ to exist, the range of the inner function must be a subset of the domain of the outer function, i.e., $R_{\mathrm{g}^{-1}} \subseteq D_{\mathrm{h}}$.

Comparing the intervals:
$[1, 4) \not\subseteq [0, 3]$

Since $R_{\mathrm{g}^{-1}} \not\subseteq D_{\mathrm{h}}$, $\mathrm{h}\mathrm{g}^{-1}$ does not exist.

1. Find the expression for $\mathrm{h}\mathrm{g}(x)$ when $1 \le x < 3$.

For $1 \le x < 3$, $\mathrm{g}(x) = 4 - x$.
The range of values for $\mathrm{g}(x)$ on this interval is $1 < \mathrm{g}(x) \le 3$.
Since these values fall within the domain $1 \le x \le 3$ for $\mathrm{h}(x)$, we use the second branch of $\mathrm{h}(x)$:
$\mathrm{h}(\mathrm{g}(x)) = (\mathrm{g}(x) - 1)^4$
$\mathrm{h}(\mathrm{g}(x)) = (4 - x - 1)^4 = (3 - x)^4$

2. Find the expression for $\mathrm{h}\mathrm{g}(x)$ when $3 \le x < 4$.

For $3 < x < 4$, $\mathrm{g}(x) = (x - 4)^2$.
The range of values is $0 < \mathrm{g}(x) < 1$.
This falls into the domain $0 \le x < 1$ for $\mathrm{h}(x)$, so we use the first branch of $\mathrm{h}(x)$:
$\mathrm{h}(\mathrm{g}(x)) = \sqrt{1 - \mathrm{g}(x)} = \sqrt{1 - (x - 4)^2}$

Check the boundary at $x = 3$:
$\mathrm{g}(3) = (3 - 4)^2 = 1$, which falls into the domain $1 \le x \le 3$ for $\mathrm{h}(x)$.
$\mathrm{h}(\mathrm{g}(3)) = \mathrm{h}(1) = (1 - 1)^4 = 0$.
Since substituting $x = 3$ into $\sqrt{1 - (x - 4)^2}$ also yields $\sqrt{1 - (-1)^2} = 0$, the single expression $\sqrt{1 - (x - 4)^2}$ is valid for the whole interval $3 \le x < 4$.

3. Combine the results to state $\mathrm{h}\mathrm{g}(x)$ in piecewise form.

$\mathrm{h}\mathrm{g}(x) = \begin{cases} (3 - x)^4 & \text{for } 1 \le x < 3, \\ \sqrt{1 - (x - 4)^2} & \text{for } 3 \le x < 4. \end{cases}$

1. Determine the range of the inner function $\mathrm{g}(x)$ over its domain $[1, 4)$.

For the first branch $1 \le x < 3$, $\mathrm{g}(x) = 4 - x$.
As $x$ goes from $1$ to $3$, $\mathrm{g}(x)$ takes values in $(1, 3]$.
For the second branch $3 \le x < 4$, $\mathrm{g}(x) = (x - 4)^2$.
As $x$ goes from $3$ to $4$, $\mathrm{g}(x)$ takes values in $(0, 1]$.
The overall range of $\mathrm{g}$ is $(1, 3] \cup (0, 1] = (0, 3]$.
(Note: This is also the set of input values that will be passed into $\mathrm{h}$.)

2. Determine the range of $\mathrm{h}$ over the input interval $(0, 3]$ to find the range of $\mathrm{h}\mathrm{g}$.

We evaluate $\mathrm{h}(y)$ for $y \in (0, 3]$.
For $y \in (0, 1)$, $\mathrm{h}(y) = \sqrt{1 - y}$, which takes values in the interval $(0, 1)$.
For $y \in [1, 3]$, $\mathrm{h}(y) = (y - 1)^4$.
The minimum occurs at $y = 1$ giving $0$, and the maximum at $y = 3$ giving $(3 - 1)^4 = 16$. This yields the interval $[0, 16]$.
The overall range of $\mathrm{h}\mathrm{g}$ is the union of these two intervals:
$R_{\mathrm{h}\mathrm{g}} = (0, 1) \cup [0, 16] = [0, 16]$.