The function $\text{f}$ is defined by $\text{f}(x) = (x+1)|x+1|, \quad \text{for } x \in \mathbb{R}, -4 < x \le 2.$

1. Express $\text{f}(x)$ without the modulus sign by splitting it into two intervals based on the sign of $x+1$.

$\text{f}(x) = \begin{cases} (x+1)(x+1) & \text{for } x+1 \ge 0 \\ (x+1)[-(x+1)] & \text{for } x+1 < 0 \end{cases}$
$\text{f}(x) = \begin{cases} (x+1)^2 & \text{for } -1 \le x \le 2 \\ -(x+1)^2 & \text{for } -4 < x < -1 \end{cases}$

2. Determine the range of $\text{f}$ for each interval to establish the domain of each piece of the inverse function.

For $-1 \le x \le 2$:
$\text{f}(-1) = 0, \quad \text{f}(2) = 9 \implies 0 \le y \le 9$
For $-4 < x < -1$:
$\text{f}(-4) = -9, \quad \text{f}(-1) = 0 \implies -9 < y < 0$

3. Find the inverse rule for the non-negative piece.

Let $y = (x+1)^2$ for $0 \le y \le 9$.
$x+1 = \pm\sqrt{y}$
Since $x \ge -1$, $x+1 \ge 0$, so we select the positive root:
$x+1 = \sqrt{y} \implies x = \sqrt{y} - 1$

4. Find the inverse rule for the negative piece.

Let $y = -(x+1)^2$ for $-9 < y < 0$.
$(x+1)^2 = -y \implies x+1 = \pm\sqrt{-y}$
Since $x < -1$, $x+1 < 0$, so we select the negative root:
$x+1 = -\sqrt{-y} \implies x = -\sqrt{-y} - 1$

5. State the final piecewise inverse function $\text{f}^{-1}$ with its full domain.

Replacing $y$ with $x$, we combine both pieces to form $\text{f}^{-1}(x)$:
$\text{f}^{-1}(x) = \begin{cases} \sqrt{x} - 1 & \text{for } 0 \le x \le 9 \\ -\sqrt{-x} - 1 & \text{for } -9 < x < 0 \end{cases}$

1. Identify the key features and end-points of $y = \text{f}(x)$.

The function $\text{f}(x) = (x+1)|x+1|$ has domain $(-4, 2]$.
At $x = -4$, $y = (-3)|-3| = -9$.
At $x = 2$, $y = 3|3| = 9$.
The end-points are $(-4, -9)$ (which must be an unfilled circle since $x > -4$) and $(2, 9)$ (which is a filled circle).
It crosses the axes at $(-1, 0)$ and $(0, 1)$.

2. Identify the key features and end-points of $y = \text{f}^{-1}(x)$.

The graph of $y = \text{f}^{-1}(x)$ is the reflection of $y = \text{f}(x)$ in the line $y = x$.
Its domain is the range of $\text{f}(x)$, which is $(-9, 9]$.
The end-points are the reflections of $\text{f}$'s end-points: $(-9, -4)$ (unfilled circle) and $(9, 2)$ (filled circle).
It crosses the axes at $(0, -1)$ and $(1, 0)$.

3. Identify the key features and end-points of $y = \text{f}\text{f}^{-1}(x)$.

By definition, $\text{f}\text{f}^{-1}(x) = x$.
The domain of this composite function is the domain of the inner function $\text{f}^{-1}(x)$, which is $(-9, 9]$.
Thus, the graph is the straight line segment $y = x$ with end-points at $(-9, -9)$ (unfilled circle) and $(9, 9)$ (filled circle).

4. Describe the combined sketch.

Sketch all three graphs on the same axes:

• Ensure $y = \text{f}(x)$ and $y = \text{f}^{-1}(x)$ are symmetrical about the line $y = x$.
• Ensure $y = \text{f}\text{f}^{-1}(x)$ overlays the line $y = x$ exactly within the domain $-9 < x \le 9$.
• Explicitly label the coordinates of the 6 determined end-points and visually distinguish the inclusive (filled) and exclusive (unfilled) bounds.

1. Relate the inequality $\text{f}(x) \le \text{f}^{-1}(x)$ to $\text{f}(x) \le x$

Since $\text{f}(x)$ is a strictly increasing function, the graphs of $y = \text{f}(x)$ and $y = \text{f}^{-1}(x)$ only intersect on the line $y = x$. Furthermore, the inequality $\text{f}(x) \le \text{f}^{-1}(x)$ is equivalent to solving $\text{f}(x) \le x$, provided $x$ is evaluated in the common domain of $\text{f}$ and $\text{f}^{-1}$.
(The common domain is $-4 < x \le 2$).

2. Solve the inequality for the interval $x \ge -1$

For $x \ge -1$:
$\text{f}(x) \le x \implies (x+1)^2 \le x$
$x^2 + 2x + 1 \le x$
$x^2 + x + 1 \le 0$
The discriminant of the quadratic is $\Delta = 1^2 - 4(1)(1) = -3 < 0$. Since the coefficient of $x^2$ is positive, the quadratic is always positive ($x^2 + x + 1 > 0$) for all real $x$. Therefore, there are no solutions in this interval.

3. Solve the inequality for the interval $-4 < x < -1$ and combine with the domain

For $-4 < x < -1$:
$\text{f}(x) \le x \implies -(x+1)^2 \le x$
$-(x^2 + 2x + 1) \le x$
$x^2 + 3x + 1 \ge 0$
Setting the quadratic to zero gives the exact roots:
$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2} = \frac{-3 \pm \sqrt{5}}{2}$
Since the parabola $y = x^2 + 3x + 1$ opens upwards, the inequality $x^2 + 3x + 1 \ge 0$ yields:
$x \le \frac{-3 - \sqrt{5}}{2} \quad \text{or} \quad x \ge \frac{-3 + \sqrt{5}}{2}$
Given the interval constraint $-4 < x < -1$, we note that $\frac{-3 + \sqrt{5}}{2} \approx -0.382$ is outside this domain, while $\frac{-3 - \sqrt{5}}{2} \approx -2.618$ lies within it.
Taking the intersection with the domain's strict lower bound $x > -4$, we get the final exact solution:
$-4 < x \le \frac{-3 - \sqrt{5}}{2}$