配方 $2x^2 - 9x + 5$。

1. Find the range of $\mathrm{f}$ to determine the domain of $\mathrm{f}^{-1}$.

From completing the square in the previous part, $\mathrm{f}(x) = 2\left(x - \frac{9}{4}\right)^2 - \frac{41}{8}$.
Since the given domain is $1 \le x \le 2$, the values of $x$ are to the left of the turning point $x = 2.25$, meaning $\mathrm{f}$ is strictly decreasing in this interval.
Maximum value $= \mathrm{f}(1) = 2(1)^2 - 9(1) + 5 = -2$
Minimum value $= \mathrm{f}(2) = 2(2)^2 - 9(2) + 5 = -5$
The range of $\mathrm{f}$ is $[-5, -2]$.
By definition, the domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$, so $\mathrm{D}_{\mathrm{f}^{-1}} = [-5, -2]$.

2. Determine the solution set by comparing the domains of $\mathrm{f}$ and $\mathrm{f}^{-1}$.

For the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ to have a solution, $x$ must be valid for both functions. That is, $x$ must lie in the intersection of their domains.
$\mathrm{D}_{\mathrm{f}} = [1, 2]$
$\mathrm{D}_{\mathrm{f}^{-1}} = [-5, -2]$
Since $[1, 2] \cap [-5, -2] = \emptyset$, there are no values of $x$ for which both $\mathrm{f}(x)$ and $\mathrm{f}^{-1}(x)$ are simultaneously defined.
Therefore, the equation has no solutions.

1. Determine the range of the inner function $\mathrm{g}$.

Given $\mathrm{g}(x) = \frac{1}{x}$ for $0 < x \le 5$, we find the range of $\mathrm{g}$ ($\mathrm{R}_\mathrm{g}$).
As $x$ approaches $0$ from the positive side, $\mathrm{g}(x) \to \infty$.
At the maximum value of the domain $x = 5$, $\mathrm{g}(5) = \frac{1}{5}$.
Because $\mathrm{g}(x)$ is continuous and decreasing for $x > 0$, the range of $\mathrm{g}$ is:
$\mathrm{R}_\mathrm{g} = \left[\frac{1}{5}, \infty\right)$

2. Determine the domain of the outer function $\mathrm{f}^{-1}$.

The domain of the inverse function $\mathrm{f}^{-1}$ ($\mathrm{D}_{\mathrm{f}^{-1}}$) is equal to the range of $\mathrm{f}$ ($\mathrm{R}_\mathrm{f}$).
Using the expression from the previous part, $\mathrm{f}(x) = 2x^2 - 9x + 5 = 2(x - 2.25)^2 - 5.125$ for $1 \le x \le 2$.
The vertex of the parabola is at $x = 2.25$. Since the domain $[1, 2]$ lies entirely to the left of the vertex, $\mathrm{f}(x)$ is strictly decreasing on this interval.
Evaluating at the endpoints:
$\mathrm{f}(1) = 2(1)^2 - 9(1) + 5 = -2$
$\mathrm{f}(2) = 2(2)^2 - 9(2) + 5 = -5$
Thus, the range of $\mathrm{f}$ is $[-5, -2]$, meaning:
$\mathrm{D}_{\mathrm{f}^{-1}} = [-5, -2]$

3. Check the condition for the existence of the composite function $\mathrm{f}^{-1}\mathrm{g}$.

For the composite function $\mathrm{f}^{-1}\mathrm{g}$ to exist, the range of the inner function must be a subset of the domain of the outer function, i.e., we require $\mathrm{R}_\mathrm{g} \subseteq \mathrm{D}_{\mathrm{f}^{-1}}$.
Comparing the two sets:
$\left[\frac{1}{5}, \infty\right) \not\subseteq [-5, -2]$
Since the condition is not met, the composite function $\mathrm{f}^{-1}\mathrm{g}$ does not exist.

1. State the condition for the existence of the composite function and find the range of the restricted function.

For the composite function $\mathrm{g}\mathrm{g}$ to exist, the range of the restricted $\mathrm{g}$ must be a subset of its domain, so $R_{\mathrm{g}} \subseteq D_{\mathrm{g}}$.

The restricted domain is $D_{\mathrm{g}} = [p, q]$. Since $p > 0$ and $\mathrm{g}(x) = \frac{1}{x}$ is a strictly decreasing function for positive $x$, the range is $R_{\mathrm{g}} = \left[\frac{1}{q}, \frac{1}{p}\right]$.

2. Set up inequalities from the subset condition to find the equation involving $p$ and $q$.

For $\left[\frac{1}{q}, \frac{1}{p}\right] \subseteq [p, q]$, the lower and upper bounds must satisfy:
$\frac{1}{q} \ge p \quad \text{and} \quad \frac{1}{p} \le q$

Since $p$ and $q$ are positive, multiplying by the denominators preserves the inequalities:
$1 \ge pq \quad \text{and} \quad 1 \le pq$

For both conditions to be true simultaneously, we must have:
$pq = 1$