We are given a function $f(x) = \frac{x+1}{\lfloor 3x-2 \rfloor}$, where $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$. (a) We need to find the domain of $f(x)$. (b) We need to discuss the continuity of $f(x)$ at $x = -1$.

1. Problem Description

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

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

(b) We need to discuss the continuity of $f(x)$ at $x = -1$.

2. Solution Steps

(a) The domain of $f(x)$ is the set of all real numbers $x$ for which the denominator is not zero.

Thus, we need to find the values of $x$ for which $\lfloor 3x-2 \rfloor \ne 0$.

$\lfloor 3x-2 \rfloor = 0$ if and only if $0 \le 3x-2 < 1$.

Adding 2 to each part of the inequality, we have $2 \le 3x < 3$.

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

Thus, the domain of $f(x)$ is the set of all real numbers $x$ such that $x \notin [\frac{2}{3}, 1)$.

In interval notation, the domain is $(-\infty, \frac{2}{3}) \cup [1, \infty)$.

(b) To discuss the continuity of $f(x)$ at $x = -1$, we need to evaluate $f(-1)$.

$f(-1) = \frac{-1+1}{\lfloor 3(-1)-2 \rfloor} = \frac{0}{\lfloor -5 \rfloor} = \frac{0}{-5} = 0$.

We need to check if $\lim_{x \to -1} f(x) = f(-1) = 0$.

For $x$ close to $-1$, say in the interval $(-1.5, -0.5)$, we have $3x-2$ between $-6.5$ and $-3.5$.

Then $\lfloor 3x-2 \rfloor$ is a constant, equal to $-6$.

$\lim_{x \to -1} f(x) = \lim_{x \to -1} \frac{x+1}{\lfloor 3x-2 \rfloor} = \lim_{x \to -1} \frac{x+1}{-5} = \frac{-1+1}{-5} = \frac{0}{-5} = 0$.

Since $\lim_{x \to -1} f(x) = f(-1) = 0$, $f(x)$ is continuous at $x = -1$.

3. Final Answer

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

(b) $f(x)$ is continuous at $x = -1$.