函数 $\text{f}$ 定义为
$\text{f} : x \mapsto 2x^2 + 5x - 3, \quad x \in \mathbb{R}, \quad x \le a.$

1. Find the $x$-coordinate of the turning point of the quadratic graph.

To find the turning point, complete the square for $\text{f}(x)$:
$\text{f}(x) = 2\left(x^2 + \frac{5}{2}x\right) - 3$
$= 2\left(x + \frac{5}{4}\right)^2 - 2\left(\frac{5}{4}\right)^2 - 3$
$= 2\left(x + \frac{5}{4}\right)^2 - \frac{49}{8}$
The turning point is a minimum at $x = -\frac{5}{4}$.

2. Deduce the greatest value of $a$ for the function to be one-to-one.

For $\text{f}^{-1}$ to exist, the function $\text{f}$ must be one-to-one.
Since the graph of $\text{f}$ is a parabola opening upwards, a restricted domain of $x \le a$ will be one-to-one if and only if it does not extend past the minimum point.
Thus, we require $a \le -\frac{5}{4}$.
The greatest value of $a$ is $-\frac{5}{4}$.

1. Use the value of $a$ from part (i) and complete the square to express $x$ in terms of $y$.

From part (i), the turning point is at $x = -\frac{5}{4}$, giving the greatest value $a = -\frac{5}{4}$.
Let $y = 2x^2 + 5x - 3$.
Completing the square:
$\begin{aligned} y &= 2\left(x^2 + \frac{5}{2}x\right) - 3 \\ &= 2\left[ \left(x + \frac{5}{4}\right)^2 - \left(\frac{5}{4}\right)^2 \right] - 3 \\ &= 2\left(x + \frac{5}{4}\right)^2 - 2\left(\frac{25}{16}\right) - 3 \\ &= 2\left(x + \frac{5}{4}\right)^2 - \frac{49}{8} \end{aligned}$
Solving for $x$:
$\begin{aligned} 2\left(x + \frac{5}{4}\right)^2 &= y + \frac{49}{8} = \frac{8y + 49}{8} \\ \left(x + \frac{5}{4}\right)^2 &= \frac{8y + 49}{16} \\ x + \frac{5}{4} &= \pm \frac{\sqrt{8y + 49}}{4} \\ x &= \frac{-5 \pm \sqrt{8y + 49}}{4} \end{aligned}$

2. Select the correct root based on the domain restriction of $\text{f}$.

The domain of $\text{f}$ is $x \le a$, which means $x \le -\frac{5}{4}$.
Because $x + \frac{5}{4} \le 0$, we must reject the positive square root and take the negative root:
$x = \frac{-5 - \sqrt{8y + 49}}{4}$

3. Determine the domain of the inverse function and express it in mapping notation.

The minimum value of $\text{f}(x)$ is $-\frac{49}{8}$, so the range of $\text{f}$ is $y \ge -\frac{49}{8}$.
The domain of $\text{f}^{-1}$ corresponds to the range of $\text{f}$, hence the domain is $x \ge -\frac{49}{8}$.
Writing this in a similar mapping form as $\text{f}$:
$\text{f}^{-1} : x \mapsto \frac{-5 - \sqrt{8x + 49}}{4}, \quad x \in \mathbb{R}, \quad x \ge -\frac{49}{8}$

1. Determine the domain restriction $a$ from part (i) and set up the equivalent equation $\text{f}(x) = x$.

To find the value of $a$ from part (i), we find the $x$-coordinate of the vertex of the parabola $\text{f}(x) = 2x^2 + 5x - 3$.
$x = -\frac{b}{2a} = -\frac{5}{2(2)} = -\frac{5}{4}$
Thus, $a = -\frac{5}{4}$, and the domain of $\text{f}$ is $x \le -\frac{5}{4}$.

Solving $\text{f}^{-1}(x) = x$ is equivalent to solving $\text{f}(x) = x$.
$2x^2 + 5x - 3 = x$
$2x^2 + 4x - 3 = 0$

2. Solve the resulting quadratic equation using the quadratic formula.

Using the quadratic formula:
$x = \frac{-4 \pm \sqrt{4^2 - 4(2)(-3)}}{2(2)}$
$x = \frac{-4 \pm \sqrt{16 + 24}}{4}$
$x = \frac{-4 \pm \sqrt{40}}{4}$
$x = \frac{-4 \pm 2\sqrt{10}}{4} = -1 \pm \frac{\sqrt{10}}{2}$

3. Reject the extraneous solution based on the domain of the function.

Since the domain of $\text{f}$ is $x \le -\frac{5}{4}$ (or $x \le -1.25$):
$x = -1 + \frac{\sqrt{10}}{2} \approx 0.581 \quad \text{(rejected)}$
$x = -1 - \frac{\sqrt{10}}{2} \approx -2.581 \quad \text{(accepted)}$

The exact solution is $x = -1 - \frac{\sqrt{10}}{2}$.

1. Apply the original function $\text{f}$ to both sides to eliminate the inverse function $\text{f}^{-1}$.

$\text{f}^{-1}(\text{g}(b)) = -3$\n$\text{g}(b) = \text{f}(-3)$

2. Evaluate $\text{f}(-3)$ to find the numerical value of $\text{g}(b)$.

Substitute $x = -3$ into $\text{f}(x) = 2x^2 + 5x - 3$:\n$\text{g}(b) = 2(-3)^2 + 5(-3) - 3$\n$\text{g}(b) = 18 - 15 - 3$\n$\text{g}(b) = 0$

3. Use the provided table to identify the value of $b$ that corresponds to an output of $0$.

From the given table, the value of $x$ that results in $\text{g}(x) = 0$ is $0$.\n$b = 0$