若对于函数 $\mathrm{f}$ 定义域内的所有 $x$，都有 $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$，则称该函数为自身反函数。 函数 $\mathrm{g}$ 定义为
$\mathrm{g} : x \mapsto \sqrt{\frac{x^2 + 2}{x^2 - 1}}, \quad x > 1.$

1. Determine the equations of the vertical and horizontal asymptotes.

To find the vertical asymptote, look for where the denominator approaches $0$ at the boundary of the domain:
$x^2 - 1 = 0 \implies x = 1 \text{ (since } x > 1)$
So the vertical asymptote is $x = 1$.

To find the horizontal asymptote, evaluate the limit as $x \to \infty$:
$\lim_{x \to \infty} \sqrt{\frac{x^2+2}{x^2-1}} = \lim_{x \to \infty} \sqrt{\frac{1+\frac{2}{x^2}}{1-\frac{1}{x^2}}} = \sqrt{1} = 1$
So the horizontal asymptote is $y = 1$.

2. Sketch the curve restricted to the given domain.

For $x > 1$, the function has no $x$-intercepts or $y$-intercepts.
The curve is strictly decreasing from infinity (near $x = 1$) towards the horizontal line $y = 1$ (as $x \to \infty$).
Draw the vertical dotted line $x=1$ and horizontal dotted line $y=1$, then draw a single smooth curve in the region $x>1, y>1$ approaching both asymptotes.

1. Set up the equation to find the inverse by letting $y = \mathrm{g}(x)$ and squaring both sides.

Let $y = \sqrt{\frac{x^2 + 2}{x^2 - 1}}$.

Squaring both sides:
$y^2 = \frac{x^2 + 2}{x^2 - 1}$
$y^2(x^2 - 1) = x^2 + 2$
$x^2 y^2 - y^2 = x^2 + 2$

2. Rearrange to make $x^2$ the subject.

$x^2 y^2 - x^2 = y^2 + 2$
$x^2(y^2 - 1) = y^2 + 2$
$x^2 = \frac{y^2 + 2}{y^2 - 1}$

3. Take the square root to make $x$ the subject, considering the domain of $x$.

Since the domain of $\mathrm{g}$ is $x > 1$, $x$ must be positive. Thus, we take the positive square root:
$x = \sqrt{\frac{y^2 + 2}{y^2 - 1}}$

4. Determine the domain of $\mathrm{g}^{-1}$ by finding the range of $\mathrm{g}$, and define $\mathrm{g}^{-1}$.

The domain of $\mathrm{g}^{-1}$ is the range of $\mathrm{g}$.
For $x > 1$, as $x \to 1^+$, $y \to \infty$, and as $x \to \infty$, $y \to 1$.
Hence, the range of $\mathrm{g}$ is $y > 1$.

Defining $\mathrm{g}^{-1}$ in a similar form:
$\mathrm{g}^{-1} : x \mapsto \sqrt{\frac{x^2 + 2}{x^2 - 1}}, \quad x > 1.$

5. Conclude that $\mathrm{g}$ is self-inverse.

Comparing the definitions of $\mathrm{g}$ and $\mathrm{g}^{-1}$, we see that:
$\mathrm{g}^{-1}(x) = \mathrm{g}(x)$
for all $x > 1$ (which is the domain of $\mathrm{g}$).
Therefore, $\mathrm{g}$ is self-inverse.

1. Show that $\mathrm{g}^2(x) = x$ and $\mathrm{g}^3(x) = \mathrm{g}(x)$.

$\mathrm{g}^2(x) = \mathrm{g}(\mathrm{g}(x)) = \sqrt{\frac{[\mathrm{g}(x)]^2 + 2}{[\mathrm{g}(x)]^2 - 1}}$
$= \sqrt{\frac{\frac{x^2 + 2}{x^2 - 1} + 2}{\frac{x^2 + 2}{x^2 - 1} - 1}}$
$= \sqrt{\frac{\frac{x^2 + 2 + 2(x^2 - 1)}{x^2 - 1}}{\frac{x^2 + 2 - (x^2 - 1)}{x^2 - 1}}}$
$= \sqrt{\frac{3x^2}{3}} = \sqrt{x^2}$
Since the domain is $x > 1$, $\sqrt{x^2} = x$. Thus, $\mathrm{g}^2(x) = x$.
(Alternatively, from part (ii), $\mathrm{g}$ is self-inverse so $\mathrm{g}^{-1}(x) = \mathrm{g}(x)$, hence $\mathrm{g}^2(x) = \mathrm{g}(\mathrm{g}^{-1}(x)) = x$.)

For $\mathrm{g}^3(x)$:
$\mathrm{g}^3(x) = \mathrm{g}(\mathrm{g}^2(x)) = \mathrm{g}(x)$

2. Simplify the powers $\mathrm{g}^{50}(x)$ and $\mathrm{g}^{51}(x)$ and set up the equation.

Since $\mathrm{g}^2(x) = x$, any even iteration yields $x$ and any odd iteration yields $\mathrm{g}(x)$.
Therefore, $\mathrm{g}^{50}(x) = x$ and $\mathrm{g}^{51}(x) = \mathrm{g}(x)$.

The equation $4 - \mathrm{g}^{50}(x) = \left[\mathrm{g}^{51}(x)\right]^2$ becomes:
$4 - x = [\mathrm{g}(x)]^2$
$4 - x = \frac{x^2 + 2}{x^2 - 1}$

3. Rearrange into a polynomial equation and solve for $x$.

$(4 - x)(x^2 - 1) = x^2 + 2$
$4x^2 - 4 - x^3 + x = x^2 + 2$
$x^3 - 3x^2 - x + 6 = 0$

By trial and error, let $P(x) = x^3 - 3x^2 - x + 6$.
$P(2) = 8 - 12 - 2 + 6 = 0$, so $(x - 2)$ is a factor.
Using polynomial division or comparing coefficients:
$x^3 - 3x^2 - x + 6 = (x - 2)(x^2 - x - 3) = 0$

So, $x = 2$ or $x^2 - x - 3 = 0$.
For the quadratic factor:
$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2} = \frac{1 \pm \sqrt{13}}{2}$

4. Check solutions against the domain of $\mathrm{g}$.

The domain of $\mathrm{g}$ is given as $x > 1$.

• $x = 2 > 1$ (Accept)
• $x = \frac{1 + \sqrt{13}}{2} \approx 2.30 > 1$ (Accept)
• $x = \frac{1 - \sqrt{13}}{2} \approx -1.30 < 1$ (Reject)

The valid values are $x = 2$ and $x = \frac{1 + \sqrt{13}}{2}$.