函数 $\text{f}$ 的定义为
$\text{f} : x \mapsto \frac{1}{x^2 - x - 6} + 2, \quad x \in \mathbb{R}, \ x \ne -2, \ x \ne 3.$

1. Evaluate the function at two distinct points to find a counterexample for the one-to-one property.

Let $x = 0$,
$\text{f}(0) = \frac{1}{0^2 - 0 - 6} + 2 = -\frac{1}{6} + 2 = \frac{11}{6}$

Let $x = 1$,
$\text{f}(1) = \frac{1}{1^2 - 1 - 6} + 2 = -\frac{1}{6} + 2 = \frac{11}{6}$

2. Conclude that the function is not one-to-one, meaning its inverse does not exist.

Since $\text{f}(0) = \text{f}(1)$ for $0 \ne 1$, the function $\text{f}$ is not a one-to-one function.

Therefore, the inverse function $\text{f}^{-1}$ does not exist.

1. Find the first derivative of $\text{f}(x)$.

Rewrite $\text{f}(x)$ using a negative index:
$\text{f}(x) = (x^2 - x - 6)^{-1} + 2$

Differentiate with respect to $x$ using the chain rule:
$\text{f}'(x) = -(x^2 - x - 6)^{-2}(2x - 1)$
$\text{f}'(x) = \frac{1 - 2x}{(x^2 - x - 6)^2}$

2. Set up the inequality for a decreasing function.

For the function to be decreasing, its gradient must be negative, so $\text{f}'(x) < 0$.
$\frac{1 - 2x}{(x^2 - x - 6)^2} < 0$

3. Solve the rational inequality.

The denominator $(x^2 - x - 6)^2$ is a perfect square, meaning $(x^2 - x - 6)^2 > 0$ for all valid $x$ in the domain (where $x \ne -2$ and $x \ne 3$).

Therefore, the sign of the derivative depends entirely on the numerator. For the fraction to be negative, we must have:
$1 - 2x < 0$
$2x > 1 \implies x > \frac{1}{2}$

4. Combine with the domain restrictions to state the final set of values.

The given domain restricts $x \ne -2$ and $x \ne 3$.
The condition $x > \frac{1}{2}$ already excludes $-2$. However, $x = 3$ is greater than $\frac{1}{2}$ and thus falls within this region, so it must still be explicitly excluded.

The set of values is therefore $x > \frac{1}{2}, x \ne 3$.
(Note: $x \ge \frac{1}{2}, x \ne 3$ is also acceptable depending on the rigorous definition of decreasing vs strictly decreasing)

1. Find an expression for $\text{gf}(x)$ by substituting $\text{f}(x)$ into $\text{g}(x)$.

$\text{gf}(x) = \text{g}\left( \frac{1}{x^2 - x - 6} + 2 \right)$
$\text{gf}(x) = 2 - \left( \frac{1}{x^2 - x - 6} + 2 \right)$
$\text{gf}(x) = -\frac{1}{x^2 - x - 6}$

2. Set up an equation to evaluate the inverse function at $\frac{1}{4}$ and solve the resulting quadratic equation.

Let $x = (\text{gf})^{-1}\left(\frac{1}{4}\right) \implies \text{gf}(x) = \frac{1}{4}$.
$-\frac{1}{x^2 - x - 6} = \frac{1}{4}$
$x^2 - x - 6 = -4$
$x^2 - x - 2 = 0$
$(x - 2)(x + 1) = 0$
$x = 2 \quad \text{or} \quad x = -1$

3. Determine the valid root by considering the domain restriction.

Since the domain of $\text{f}$ is restricted to $x \le \frac{1}{2}$, the domain of $\text{gf}$ is also $x \le \frac{1}{2}$.
We reject $x = 2$.
$\therefore (\text{gf})^{-1}\left(\frac{1}{4}\right) = -1$