The function $\mathrm{f}$ is defined by
$\mathrm{f} : x \mapsto \ln x^2 - (\ln x)^2, \quad x > 0.$
Sketch the graph of $y = \mathrm{f}(x)$, indicating clearly all points of intersection with the axes. Explain why the inverse function $\mathrm{f}^{-1}$ does not exist. The function $\mathrm{h}$ is defined by
$\mathrm{h} : x \mapsto \ln x^2 - (\ln x)^2, \quad 0 < x \le a.$
State the greatest possible value of $a$ for which the inverse function $\mathrm{h}^{-1}$ exists. Find the inverse function $\mathrm{h}^{-1}$ in similar form. The function $\mathrm{g}$ is defined by
$\mathrm{g} : x \mapsto 3\mathrm{e}^x, \quad x \in \mathbb{R}.$
Find the range of the composite function $\mathrm{gh}$.

1. Find the coordinates of the intersections with the axes.

For $x > 0$, we have $y = \ln x^2 - (\ln x)^2 = 2 \ln x - (\ln x)^2 = (\ln x)(2 - \ln x)$.

To find the $x$-intercepts, set $y = 0$:
$\ln x = 0 \quad \text{or} \quad \ln x = 2$
$x = 1 \quad \text{or} \quad x = \mathrm{e}^2$
The points of intersection with the $x$-axis are $(1, 0)$ and $(\mathrm{e}^2, 0)$.
There is no $y$-intercept since the domain is $x > 0$.

2. Sketch the graph of $y = \mathrm{f}(x)$ using key features.

Using a graphing calculator or by completing the square to find the maximum point $(\mathrm{e}, 1)$ and observing the vertical asymptote at $x = 0$ (the $y$-axis):

(The student draws a smooth curve starting from the bottom of the $y$-axis, passing through $(1,0)$, reaching a maximum at $(\mathrm{e},1)$, and passing down through $(\mathrm{e}^2, 0)$ towards $-\infty$.)

3. Explain why the inverse function does not exist.

From the graph, any horizontal line $y = k$ for $0 \le k < 1$ will intersect the curve at two distinct points.
For instance, $\mathrm{f}(1) = \mathrm{f}(\mathrm{e}^2) = 0$.
Because the horizontal line test fails, $\mathrm{f}$ is not a one-to-one function.
Therefore, the inverse function $\mathrm{f}^{-1}$ does not exist.

1. Determine the greatest value of $a$ for which the function is one-to-one by completing the square.

$\mathrm{h}(x) = \ln x^2 - (\ln x)^2$
$= 2\ln x - (\ln x)^2$
$= -((\ln x)^2 - 2\ln x)$
$= -((\ln x - 1)^2 - 1)$
$= 1 - (\ln x - 1)^2$

The graph has a maximum turning point when $\ln x - 1 = 0 \implies x = \mathrm{e}$.
For the inverse function $\mathrm{h}^{-1}$ to exist, $\mathrm{h}$ must be a one-to-one function.
Thus, the domain $0 < x \le a$ must be restricted to the strictly increasing portion of the curve up to the turning point.
The greatest possible value of $a$ is $\mathrm{e}$.

2. Express $x$ in terms of $y$ to find the inverse function.

Let $y = \mathrm{h}(x)$.
$y = 1 - (\ln x - 1)^2$
$(\ln x - 1)^2 = 1 - y$
$\ln x - 1 = \pm\sqrt{1 - y}$

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

Since $0 < x \le \mathrm{e}$, we have $\ln x \le 1 \implies \ln x - 1 \le 0$.
Therefore, we must reject the positive root and choose the negative sign:
$\ln x - 1 = -\sqrt{1 - y}$
$\ln x = 1 - \sqrt{1 - y}$
$x = \mathrm{e}^{1 - \sqrt{1 - y}}$

4. State the inverse function in similar form, including its domain.

The domain of $\mathrm{h}^{-1}$ is the range of $\mathrm{h}$.
For $0 < x \le \mathrm{e}$, the function values increase from $-\infty$ up to a maximum of $\mathrm{h}(\mathrm{e}) = 1$.
Thus, the range of $\mathrm{h}$ is $y \le 1$, meaning the domain of $\mathrm{h}^{-1}$ is $x \le 1$.

In similar form:
$\mathrm{h}^{-1} : x \mapsto \mathrm{e}^{1 - \sqrt{1 - x}}, \quad x \le 1$

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

From part (b), the domain of $\mathrm{h}$ is $0 < x \le \mathrm{e}$.
By completing the square for the expression, we have:
$\mathrm{h}(x) = 2\ln x - (\ln x)^2 = 1 - (\ln x - 1)^2$
For $0 < x \le \mathrm{e}$, we have $\ln x \le 1 \implies \ln x - 1 \le 0$.
The maximum value of $\mathrm{h}(x)$ occurs at $x = \mathrm{e}$, giving $\mathrm{h}(\mathrm{e}) = 1$.
As $x \to 0^+$, $\ln x \to -\infty$, so $\mathrm{h}(x) \to -\infty$.
Thus, the range of $\mathrm{h}$ is $R_{\mathrm{h}} = (-\infty, 1]$.

2. Find the range of the composite function $\mathrm{gh}$ by evaluating $\mathrm{g}$ over $R_{\mathrm{h}}$.

The range of $\mathrm{gh}$ is the image of $R_{\mathrm{h}}$ under $\mathrm{g}$.
$\mathrm{g}(x) = 3\mathrm{e}^x$ is a strictly increasing function.
For the upper bound, substituting the maximum value of $R_{\mathrm{h}}$:
$\mathrm{g}(1) = 3\mathrm{e}^1 = 3\mathrm{e}$
For the lower bound, taking the limit as the input approaches $-\infty$:
As $y \to -\infty$, $\mathrm{g}(y) = 3\mathrm{e}^y \to 0$.
Therefore, the range of $\mathrm{gh}$ is $(0, 3\mathrm{e}]$.