已知 $f(x) = \begin{cases} \tfrac{1}{2}x^2, & \text{for } 1 \le x < 2, \\ \tfrac{1}{4}(-3x+14), & \text{for } 2 \le x < 4, \end{cases}$ 且对于所有实数 $x$，都有 $f(x) = f(x+3)$。

1. Relate $f(0)$ to a value within the defined domain using the given periodic condition.

Since $f(x) = f(x+3)$ for all real values of $x$, we can substitute $x = 0$:
$f(0) = f(0+3) = f(3)$

2. Evaluate $f(3)$ using the piecewise definition of the function.

The value $x = 3$ falls in the interval $2 \le x < 4$, so we use $f(x) = \frac{1}{4}(-3x+14)$.
$f(3) = \frac{1}{4}(-3(3) + 14)$
$f(3) = \frac{1}{4}(5) = \frac{5}{4}$

1. Evaluate the key points and determine the shapes for the base domain $1 \le x < 4$.

For $1 \le x < 2$, the function is $f(x) = \frac{1}{2}x^2$, which is a parabola opening upwards.
At $x=1$, $f(1) = \frac{1}{2}(1)^2 = 0.5$.
As $x \to 2^-$, $f(2) = \frac{1}{2}(2)^2 = 2$.

For $2 \le x < 4$, the function is $f(x) = \frac{1}{4}(-3x+14)$, which is a straight line with a negative gradient.
At $x=2$, $f(2) = \frac{1}{4}(-6+14) = 2$.
As $x \to 4^-$, $f(4) = \frac{1}{4}(-12+14) = 0.5$.

2. Use the periodic property $f(x) = f(x+3)$ to deduce the graph for intervals $0 \le x < 1$ and $4 \le x \le 5$.

The relationship $f(x) = f(x+3)$ indicates the function is periodic with period 3.
For $0 \le x < 1$, the graph is a translation of the segment for $3 \le x < 4$ by 3 units to the left.
Using part (a), $f(0) = f(3) = \frac{1}{4}(-9+14) = 1.25$.
Thus, for $0 \le x < 1$, it is a straight line segment connecting $(0, 1.25)$ to $(1, 0.5)$.

For $4 \le x \le 5$, the graph is a translation of the segment for $1 \le x \le 2$ by 3 units to the right.
At $x=5$, $f(5) = f(2) = 2$.
Thus, for $4 \le x \le 5$, it is an upward-curving parabolic arc connecting $(4, 0.5)$ to $(5, 2)$.

3. Synthesise the information to construct the final sketch.

Sketch the pieces sequentially from $x=0$ to $x=5$ on a single set of axes, ensuring continuity at the boundaries:

• Straight line from $(0, 1.25)$ to $(1, 0.5)$
• Parabolic curve from $(1, 0.5)$ to $(2, 2)$
• Straight line from $(2, 2)$ to $(4, 0.5)$
• Parabolic curve from $(4, 0.5)$ to $(5, 2)$

Ensure to plot and label all critical points: $(0, 1.25)$, $(1, 0.5)$, $(2, 2)$, $(4, 0.5)$, and $(5, 2)$.

1. Sketch the graph of $y = g(x)$ on the given domain.

$g(x) = \frac{4}{3}|x-2|$ is a V-shaped graph with its vertex at $(2, 0)$.
On the domain $0 \le x \le 5$, it passes through the $y$-axis at $(0, \frac{8}{3})$ and ends at $(5, 4)$. Draw this graph on the same axes as $y = f(x)$.

2. Identify the regions where $f(x) > g(x)$ using the graphs.

From the sketch, the graph of $y=f(x)$ is above $y=g(x)$ exactly between their two points of intersection.
The first intersection lies in the interval $1 \le x < 2$, where $f(x) = \frac{1}{2}x^2$ and $g(x) = -\frac{4}{3}(x-2)$.
The second intersection lies in the interval $2 \le x < 4$, where $f(x) = \frac{1}{4}(-3x+14)$ and $g(x) = \frac{4}{3}(x-2)$.

3. Find the $x$-coordinate of the first point of intersection.

$\frac{1}{2}x^2 = -\frac{4}{3}(x-2)$
$3x^2 = -8(x-2)$
$3x^2 + 8x - 16 = 0$
$(3x - 4)(x + 4) = 0$
Since $x > 0$ for this region, we take $x = \frac{4}{3}$.

4. Find the $x$-coordinate of the second point of intersection.

$\frac{1}{4}(-3x+14) = \frac{4}{3}(x-2)$
$-9x + 42 = 16(x-2)$
$-9x + 42 = 16x - 32$
$25x = 74$
$x = \frac{74}{25} \text{ (or } 2.96)$

5. State the final solution to the inequality.

Since $f(x) > g(x)$ between these intersection points, the solution is:
$\frac{4}{3} < x < \frac{74}{25}$

1. Determine the range of $f$ over all real values of $x$.

From the definition of $f(x)$ (or observing the graph from part (b)):
For $1 \le x < 2$, $f(x) = \frac{1}{2}x^2$ is strictly increasing, so $\frac{1}{2} \le f(x) < 2$.
For $2 \le x < 4$, $f(x) = \frac{1}{4}(-3x+14)$ is strictly decreasing, so $\frac{1}{2} < f(x) \le 2$.
Combining these, the range of $f$ on the interval $[1, 4)$ is $\left[\frac{1}{2}, 2\right]$.
Since $f(x) = f(x+3)$ for all real $x$, $f$ is periodic with period $3$. The interval $[1, 4)$ covers exactly one period, so the range of $f$ for all $x \in \mathbb{R}$ is $R_f = \left[\frac{1}{2}, 2\right]$.

2. Explain why the composite function $hf$ exists.

The domain of $h$ is given as $D_h = [0, 2]$.
For the composite function $hf$ to exist, the range of the inner function $f$ must be a subset of the domain of the outer function $h$.
Since $R_f = \left[\frac{1}{2}, 2\right]$ and $\left[\frac{1}{2}, 2\right] \subseteq [0, 2]$, we have $R_f \subseteq D_h$.
Therefore, the composite function $hf$ exists.

3. Use differentiation to find the stationary point of $h(x)$ on the interval $x \in \left[\frac{1}{2}, 2\right]$.

The range of $hf$ is the set of values of $h(x)$ for $x \in R_f = \left[\frac{1}{2}, 2\right]$.
To find the maximum and minimum values of $h(x)$ on this interval, we find its stationary points.
$h(x) = \sqrt{3}\sin(\pi x) + \cos(\pi x) + 1$
$h'(x) = \pi\sqrt{3}\cos(\pi x) - \pi\sin(\pi x)$
Setting $h'(x) = 0$:
$\pi\sin(\pi x) = \pi\sqrt{3}\cos(\pi x) \implies \tan(\pi x) = \sqrt{3}$
For $x \in \left[\frac{1}{2}, 2\right]$, the angle $\pi x \in \left[\frac{\pi}{2}, 2\pi\right]$.
The only solution in this interval is $\pi x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$, which gives the stationary point $x = \frac{4}{3}$.

4. Evaluate $h(x)$ at the stationary point and endpoints to determine the exact range of $hf$.

We evaluate $h(x)$ at the stationary point $x = \frac{4}{3}$ and the endpoints $x = \frac{1}{2}$ and $x = 2$:

• At $x = \frac{4}{3}$:
  $h\left(\frac{4}{3}\right) = \sqrt{3}\sin\left(\frac{4\pi}{3}\right) + \cos\left(\frac{4\pi}{3}\right) + 1 = \sqrt{3}\left(-\frac{\sqrt{3}}{2}\right) - \frac{1}{2} + 1 = -1$
• At the left endpoint $x = \frac{1}{2}$:
  $h\left(\frac{1}{2}\right) = \sqrt{3}\sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) + 1 = \sqrt{3}(1) + 0 + 1 = \sqrt{3} + 1$
• At the right endpoint $x = 2$:
  $h(2) = \sqrt{3}\sin(2\pi) + \cos(2\pi) + 1 = 0 + 1 + 1 = 2$
  Comparing these values, the absolute minimum is $-1$ and the absolute maximum is $\sqrt{3} + 1$.
  Thus, the range of $hf$ is $[-1, \sqrt{3} + 1]$.