曲线 $C$ 的方程为 $y = ax^2 + bx + c$，其中 $a$，$b$ 和 $c$ 是常数。 已知 $C$ 通过点 $(-2, 17)$，$\left(\frac{1}{2}, \frac{3}{4}\right)$ 和 $(5, 3)$。

1. Substitute the given points into the equation to formulate a system of linear equations.

Since the curve $y = ax^2 + bx + c$ passes through $(-2, 17)$, $\left(\frac{1}{2}, \frac{3}{4}\right)$, and $(5, 3)$, we substitute these coordinates into the equation to obtain:

For $(-2, 17)$:
$a(-2)^2 + b(-2) + c = 17$
$4a - 2b + c = 17 \quad \text{--- (1)}$

For $\left(\frac{1}{2}, \frac{3}{4}\right)$:
$a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + c = \frac{3}{4}$
$0.25a + 0.5b + c = 0.75 \quad \text{--- (2)}$

For $(5, 3)$:
$a(5)^2 + b(5) + c = 3$
$25a + 5b + c = 3 \quad \text{--- (3)}$

2. Solve the system of equations and write the final equation of the curve.

Solving the system of 3 linear equations (typically using a Graphing Calculator):
$a = 1$
$b = -5$
$c = 3$

Thus, the equation of the curve $C$ is:
$y = x^2 - 5x + 3$

1. State the function from part (a) and set up the equation to find the inverse.

From part (a), solving the simultaneous equations with the given points yields $a = 1$, $b = -5$, and $c = 3$.
Let $y = \text{f}(x) = x^2 - 5x + 3$.

2. Complete the square to make $x$ the subject of the formula.

$y = x^2 - 5x + 3$
$y = \left(x - \frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2 + 3$
$y = \left(x - \frac{5}{2}\right)^2 - \frac{13}{4}$
$\left(x - \frac{5}{2}\right)^2 = y + \frac{13}{4}$
$x - \frac{5}{2} = \pm \sqrt{y + \frac{13}{4}}$

3. Choose the correct root based on the given domain.

Since the domain of $\text{f}$ is restricted to $x \le 0$, we have $x - \frac{5}{2} \le -\frac{5}{2}$.
Therefore, we must reject the positive root and take the negative root:
$x = \frac{5}{2} - \sqrt{y + \frac{13}{4}}$
$x = \frac{5 - \sqrt{4y + 13}}{2}$

4. State the inverse function in terms of $x$.

Replacing $y$ with $x$:
$\text{f}^{-1}(x) = \frac{5 - \sqrt{4x + 13}}{2}$

1. Identify the shape and key points for the first part of the base interval, $0 \le x < 2$.

For $0 \le x < 2$, $\text{g}(x) = 3(x-1)^2$.
This is a parabola opening upwards with a minimum vertex at $(1, 0)$.
At $x = 0$, $\text{g}(0) = 3(0-1)^2 = 3$.
As $x \to 2$, $\text{g}(x) \to 3(2-1)^2 = 3$.
So the curve connects $(0, 3)$ to $(2, 3)$, dipping to $(1, 0)$.

2. Identify the shape and key points for the second part of the base interval, $2 \le x < 6$.

For $2 \le x < 6$, $\text{g}(x) = 5 - |x-4|$.
This is an inverted V-shape modulus graph.
The maximum vertex occurs when $x-4 = 0 \implies x=4$, giving $\text{g}(4) = 5$.
At $x = 2$, $\text{g}(2) = 5 - |2-4| = 3$.
As $x \to 6$, $\text{g}(x) \to 5 - |6-4| = 3$.
So the straight line segments connect $(2, 3)$ to $(4, 5)$, and $(4, 5)$ down to $(6, 3)$.

3. Use the periodic property to sketch the domain $-4 \le x < 0$.

We are given $\text{g}(x) = \text{g}(x+6)$. This means the graph repeats every $6$ units.
The interval $-4 \le x < 0$ corresponds to shifting the interval $2 \le x < 6$ left by 6 units.
The inverted V-shape is repeated with a maximum vertex at $(-2, 5)$.
The endpoints of this segment will be at $x = -4$ and $x = 0$.
At $x = -4$, $\text{g}(-4) = \text{g}(-4+6) = \text{g}(2) = 3$. Since $-4 \le x < 9$, the point $(-4, 3)$ is included as a solid dot.
The segment connects $(-4, 3)$ to $(-2, 5)$ and down to $(0, 3)$.

4. Use the periodic property to sketch the domain $6 \le x < 9$.

The interval $6 \le x < 8$ corresponds to shifting the parabola from $0 \le x < 2$ right by 6 units.
This yields a U-shape connecting $(6, 3)$ to $(8, 3)$, with a minimum vertex at $(7, 0)$.
The interval $8 \le x < 9$ corresponds to shifting the first half of the inverted V-shape (from $2 \le x < 3$) right by 6 units.
At $x = 9$, $\text{g}(9) = \text{g}(3) = 5 - |3-4| = 4$.
This gives a straight line connecting $(8, 3)$ to $(9, 4)$.
Because the domain is strictly $x < 9$, we draw a hollow circle at $(9, 4)$.

1. Find the range of the inner function $\text{h}$.

$-1 \le \cos x \le 1 \text{ for } 0 \le x \le \pi$
$-\frac{3}{2} \le \frac{3}{2}\cos x \le \frac{3}{2}$
$0 \le \frac{3}{2} + \frac{3}{2}\cos x \le 3$
Therefore, the range of $\text{h}$, denoted as $R_h$, is $[0, 3]$.

2. Prove the existence of the composite function $\text{gh}$.

The domain of $\text{g}$ is $D_g = [0, 6)$.
Since $R_h = [0, 3]$ and $[0, 3] \subseteq [0, 6)$, we have $R_h \subseteq D_g$.
Because the range of the inner function $\text{h}$ is a subset of the domain of the outer function $\text{g}$, the composite function $\text{gh}$ exists.

3. Find the range of $\text{gh}$ by evaluating $\text{g}(x)$ over the interval corresponding to $R_h$.

The range of $\text{gh}$ is the set of values $\text{g}(x)$ takes for $x \in R_h = [0, 3]$. We split this across $\text{g}$'s piecewise definitions.

For $0 \le x < 2$:
$\text{g}(x) = 3(x-1)^2$
The minimum value occurs at the turning point $x = 1$, giving $\text{g}(1) = 0$.
The maximum value is at $x=0$, where $\text{g}(0) = 3$, and as $x \to 2$, $\text{g}(x) \to 3$.
So, the range for this part is $[0, 3]$.

For $2 \le x \le 3$:
$\text{g}(x) = 5 - |x-4|$
Since $x \le 3$, $x-4 < 0$, so $|x-4| = -(x-4) = 4-x$.
$\text{g}(x) = 5 - (4-x) = x + 1$
This is a strictly increasing linear function. Its range is from $\text{g}(2) = 3$ to $\text{g}(3) = 4$.
So, the range for this part is $[3, 4]$.

Combining both parts, the range of $\text{gh}$ is $[0, 3] \cup [3, 4] = [0, 4]$.