The problem consists of several sub-problems. 1.1. Consider the function $f(x) = \begin{cases} 1 - \sqrt{-x} - 1 & \text{if } x \le 0 \\ \frac{1}{\lfloor 2x \rfloor - 3} & \text{if } x > 0 \end{cases}$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. a) Find the domain $D_f$ of $f(x)$. b) Find the range $R_f$ of $f(x)$. c) Discuss the continuity of $f(x)$ at $a = 1$. 1.2. For the function $g(x) = \begin{cases} 1 + x^2 & \text{if } x \le 0 \\ 2 - x & \text{if } 0 < x \le 2 \\ (x - 1)^2 & \text{if } x > 2 \end{cases}$ Discuss its continuity at $a = 0$. Mention any case of one-sided continuity or not. 1.3. Show that there is a number that is exactly 1 more than its cube, and use your calculator to find an interval of length 0.01 that contains such a number.

1. Problem Description

The problem consists of several sub-problems.

1.

1. Consider the function

$f(x) = \begin{cases} 1 - \sqrt{-x} - 1 & \text{if } x \le 0 \\ \frac{1}{\lfloor 2x \rfloor - 3} & \text{if } x > 0 \end{cases}$

where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.

a) Find the domain $D_f$ of $f(x)$.

b) Find the range $R_f$ of $f(x)$.

c) Discuss the continuity of $f(x)$ at $a = 1$.

1.

2. For the function

$g(x) = \begin{cases} 1 + x^2 & \text{if } x \le 0 \\ 2 - x & \text{if } 0 < x \le 2 \\ (x - 1)^2 & \text{if } x > 2 \end{cases}$

Discuss its continuity at $a = 0$. Mention any case of one-sided continuity or not.

1.

3. Show that there is a number that is exactly 1 more than its cube, and use your calculator to find an interval of length 0.01 that contains such a number.

2. Solution Steps

1.

1. a) Domain of $f(x)$:

For $x \le 0$, we require $-x \ge 0$, which is always true.

For $x > 0$, we require $\lfloor 2x \rfloor - 3 \ne 0$, so $\lfloor 2x \rfloor \ne 3$.

This means $3 \le 2x < 4$ is forbidden. Thus $1.5 \le x < 2$ is not allowed.

The domain is therefore $(-\infty, 0] \cup (0, 1.5) \cup [2, \infty)$.

1.

1. b) Range of $f(x)$:

For $x \le 0$, $f(x) = 1 - \sqrt{-x} - 1 = -\sqrt{-x} \le 0$.

For $x > 0$, $f(x) = \frac{1}{\lfloor 2x \rfloor - 3}$.

If $0 < x < 1.5$, $\lfloor 2x \rfloor$ can take values $0, 1, 2$.

So $\lfloor 2x \rfloor - 3$ can take values $-3, -2, -1$.

So $f(x)$ can take values $-\frac{1}{3}, -\frac{1}{2}, -1$.

If $x \ge 2$, $\lfloor 2x \rfloor - 3$ can take any integer value greater than or equal to

1. So $f(x)$ can take any value $\frac{1}{n}$ for $n \in \{1, 2, 3, ...\}$. The values $1, 1/2, 1/3, 1/4, ...$ are possible.

The range is $(-\infty, 0] \cup \{-1, -1/2, -1/3\} \cup (0, 1]$.

1.

1. c) Continuity of $f(x)$ at $a = 1$:

$f(1) = \frac{1}{\lfloor 2(1) \rfloor - 3} = \frac{1}{2 - 3} = -1$.

$\lim_{x \to 1^-} f(x) = \frac{1}{\lfloor 2x \rfloor - 3} = \frac{1}{\lfloor 2(1^-) \rfloor - 3} = \frac{1}{1 - 3} = -\frac{1}{2}$.

$\lim_{x \to 1^+} f(x) = \frac{1}{\lfloor 2x \rfloor - 3} = \frac{1}{\lfloor 2(1^+) \rfloor - 3} = \frac{1}{2 - 3} = -1$.

Since $\lim_{x \to 1^-} f(x) \ne \lim_{x \to 1^+} f(x)$, $f(x)$ is discontinuous at $x = 1$.

1.

2. Continuity of $g(x)$ at $a = 0$:

$g(0) = 1 + 0^2 = 1$.

$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (1 + x^2) = 1 + 0 = 1$.

$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (2 - x) = 2 - 0 = 2$.

Since $\lim_{x \to 0^-} g(x) \ne \lim_{x \to 0^+} g(x)$, $g(x)$ is discontinuous at $x = 0$.

We have left-hand continuity at $x = 0$, since $g(0) = \lim_{x \to 0^-} g(x) = 1$.

1.

3. Find a number that is exactly 1 more than its cube:

Let $x$ be the number. Then $x = x^3 + 1$, or $x^3 - x + 1 = 0$.

Let $h(x) = x^3 - x + 1$. We want to find $x$ such that $h(x) = 0$.

$h(-2) = -8 + 2 + 1 = -5$.

$h(-1) = -1 + 1 + 1 = 1$.

Since $h(-2) < 0$ and $h(-1) > 0$, there is a root in the interval $(-2, -1)$.

$h(-1.5) = -3.375 + 1.5 + 1 = -0.875$.

$h(-1.2) = -1.728 + 1.2 + 1 = 0.472$.

The root is between $-1.5$ and $-1.2$.

$h(-1.3) = -2.197 + 1.3 + 1 = 0.103$.

$h(-1.31) = -2.248091 + 1.31 + 1 = 0.061909$.

$h(-1.32) = -2.300584 + 1.32 + 1 = 0.019416$.

$h(-1.33) = -2.353737 + 1.33 + 1 = -0.023737$.

Since $h(-1.32) > 0$ and $h(-1.33) < 0$, the root is between $-1.33$ and $-1.32$.

Thus the interval $(-1.33, -1.32)$ is an interval of length 0.01 that contains such a number.

3. Final Answer

1.1 a) $(-\infty, 0] \cup (0, 1.5) \cup [2, \infty)$

1.1 b) $(-\infty, 0] \cup \{-1, -1/2, -1/3\} \cup (0, 1]$

1.1 c) $f(x)$ is discontinuous at $x = 1$.

1.2 $g(x)$ is discontinuous at $x = 0$. It is left-hand continuous at $x = 0$.

1.3 $(-1.33, -1.32)$