We are given a function $f(x) = \frac{1}{\lfloor 3x - 2 \rfloor}$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. a) We need to find the domain $D_f$ of the function $f(x)$. b) We need to discuss the continuity of $f(x)$ at $x = 0$. The function is $f(x)$ and not $g(x)$. I will assume that it is $f(x)$ not $g(x)$

1. Problem Description

We are given a function $f(x) = \frac{1}{\lfloor 3x - 2 \rfloor}$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

a) We need to find the domain $D_f$ of the function $f(x)$.

b) We need to discuss the continuity of $f(x)$ at $x = 0$. The function is $f(x)$ and not $g(x)$. I will assume that it is $f(x)$ not $g(x)$

2. Solution Steps

a) Finding the domain of $f(x)$:

The domain of $f(x)$ consists of all real numbers $x$ for which the function is defined. Since we have a fraction, the denominator cannot be zero. Therefore, we require that $\lfloor 3x - 2 \rfloor \neq 0$.

This means that $0 \le 3x - 2 < 1$ must be false. Thus, either $3x - 2 < 0$ or $3x - 2 \ge 1$.

$3x - 2 < 0$ implies $3x < 2$, so $x < \frac{2}{3}$.

$3x - 2 \ge 1$ implies $3x \ge 3$, so $x \ge 1$.

We want to find the values of $x$ for which $\lfloor 3x - 2 \rfloor = 0$.

This occurs when $0 \le 3x - 2 < 1$.

Adding 2 to all sides, we get $2 \le 3x < 3$.

Dividing by 3, we get $\frac{2}{3} \le x < 1$.

So, we need to exclude the interval $[\frac{2}{3}, 1)$ from the real numbers.

The domain of $f(x)$ is therefore $(-\infty, \frac{2}{3}) \cup [1, \infty)$.

b) Discussing the continuity of $f(x)$ at $x = 0$:

To check the continuity of $f(x)$ at $x = 0$, we need to evaluate $\lim_{x \to 0} f(x)$ and compare it with $f(0)$.

First, let's find $f(0)$:

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

Now, let's find the limit as $x$ approaches 0:

Since $f(x) = \frac{1}{\lfloor 3x - 2 \rfloor}$, we consider values of $x$ close to

0. For $x$ close to 0, $3x - 2$ is close to -

2. When $x$ is slightly greater than 0, $3x - 2$ is slightly greater than -2, so $\lfloor 3x - 2 \rfloor = -2$.

When $x$ is slightly less than 0, $3x - 2$ is slightly less than -2, so $\lfloor 3x - 2 \rfloor = -3$.

Therefore, we need to check the left and right limits.

Right-hand limit:

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{\lfloor 3x - 2 \rfloor} = \frac{1}{\lfloor -2 + \epsilon \rfloor} = \frac{1}{-2} = -\frac{1}{2}$ for small $\epsilon > 0$.

Left-hand limit:

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{\lfloor 3x - 2 \rfloor} = \frac{1}{\lfloor -2 - \epsilon \rfloor} = \frac{1}{-3} = -\frac{1}{3}$ for small $\epsilon > 0$.

Since $\lim_{x \to 0^+} f(x) \neq \lim_{x \to 0^-} f(x)$, the limit $\lim_{x \to 0} f(x)$ does not exist.

Therefore, $f(x)$ is discontinuous at $x = 0$.

3. Final Answer

a) The domain of $f(x)$ is $(-\infty, \frac{2}{3}) \cup [1, \infty)$.

b) $f(x)$ is discontinuous at $x = 0$.