## 1. Problem Description

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1. Problem Description

The problem consists of two parts.

Part 1.1 involves the function $f(x) = \sqrt{1 - \frac{x+13}{x^2 + 4x + 3}}$.

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

(b) Discuss the continuity of $f(x)$ at $a=2$, specifying any case of one-sided continuity.

Part 1.2 involves the function $g(x) = \frac{\sqrt{2x+3} - 1}{\lfloor 4x^2 - 36 \rfloor}$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.

(a) Find the domain of $g(x)$.

(b) Discuss the continuity of $g(x)$ at $a = -\frac{3}{2}$, specifying any case of one-sided continuity.

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2. Solution Steps

### 1.1 (a) Finding the domain of f(x)

To find the domain of $f(x) = \sqrt{1 - \frac{x+13}{x^2 + 4x + 3}}$, we need to ensure that the expression inside the square root is non-negative and that the denominator is not zero. Thus, we need to satisfy the following conditions:

1. $1 - \frac{x+13}{x^2 + 4x + 3} \ge 0$

2. $x^2 + 4x + 3 \ne 0$

First, let's analyze the denominator:

$x^2 + 4x + 3 = (x+1)(x+3)$. So, $x \ne -1$ and $x \ne -3$.

Now, let's solve the inequality:

$1 - \frac{x+13}{(x+1)(x+3)} \ge 0$

$\frac{(x+1)(x+3) - (x+13)}{(x+1)(x+3)} \ge 0$

$\frac{x^2 + 4x + 3 - x - 13}{(x+1)(x+3)} \ge 0$

$\frac{x^2 + 3x - 10}{(x+1)(x+3)} \ge 0$

$\frac{(x+5)(x-2)}{(x+1)(x+3)} \ge 0$

We analyze the sign of this expression by considering the intervals determined by the critical points $-5, -3, -1, 2$.

- $x < -5$: $(+)/(+)(+) = +$

- $-5 < x < -3$: $(-)/(+)(+) = -$

- $-3 < x < -1$: $(-)/(-)(+) = +$

- $-1 < x < 2$: $(-)/(-)(-) = -$

- $x > 2$: $(+)/(-)(-) = +$

We want the expression to be greater than or equal to zero.

So, we have: $x \le -5$ or $-3 < x < -1$ or $x \ge 2$.

Since the expression can be equal to zero, we include $x = -5$ and $x = 2$. We must exclude $x = -3$ and $x = -1$ as they make the denominator zero.

Thus, the domain is $(-\infty, -5] \cup (-3, -1) \cup [2, \infty)$.

### 1.1 (b) Discussing the continuity of f(x) at a=2

The function $f(x) = \sqrt{1 - \frac{x+13}{x^2 + 4x + 3}}$ is continuous at $x=2$ if $f(2)$ exists and $\lim_{x \to 2} f(x) = f(2)$.

$f(2) = \sqrt{1 - \frac{2+13}{2^2 + 4(2) + 3}} = \sqrt{1 - \frac{15}{4+8+3}} = \sqrt{1 - \frac{15}{15}} = \sqrt{1-1} = 0$. So $f(2)$ exists.

Since $f(x) = \sqrt{\frac{(x+5)(x-2)}{(x+1)(x+3)}}$, and $x > 2$, we consider $\lim_{x \to 2^+} f(x)$. As $x$ approaches $2$ from the right, $x-2$ approaches $0$ from the right. Since all the other factors are non-zero in a neighborhood of $2$, we have

$\lim_{x \to 2^+} \sqrt{\frac{(x+5)(x-2)}{(x+1)(x+3)}} = 0$.

Since the domain of $f(x)$ includes an interval to the right of $x=2$, we can evaluate the right-hand limit.

Since $f(2) = 0$ and $\lim_{x \to 2^+} f(x) = 0$, we can say that $f(x)$ is right-continuous at $x=2$.

### 1.2 (a) Finding the domain of g(x)

To find the domain of $g(x) = \frac{\sqrt{2x+3} - 1}{\lfloor 4x^2 - 36 \rfloor}$, we need to ensure that:

1. $2x+3 \ge 0$, so $x \ge -\frac{3}{2}$.

2. $\lfloor 4x^2 - 36 \rfloor \ne 0$.

From the second condition, we need to find the values of $x$ such that $0 \le 4x^2 - 36 < 1$ is not true. In other words, we need $\lfloor 4x^2 - 36 \rfloor \ne 0$, so $4x^2 - 36$ cannot lie in the interval $[0,1)$.

$0 \le 4x^2 - 36 < 1$

$36 \le 4x^2 < 37$

$9 \le x^2 < \frac{37}{4}$

$3 \le |x| < \frac{\sqrt{37}}{2}$

So we have two intervals to exclude: $3 \le x < \frac{\sqrt{37}}{2}$ and $-\frac{\sqrt{37}}{2} < x \le -3$. Since $x \ge -\frac{3}{2}$, we only consider the interval $3 \le x < \frac{\sqrt{37}}{2}$.

Since $\sqrt{37} \approx 6.08$, $\frac{\sqrt{37}}{2} \approx 3.04$. Thus, we must exclude the interval $[3, \frac{\sqrt{37}}{2})$.

Also, the numerator must be real. $\sqrt{2x+3}-1$ exists for $x \ge -\frac{3}{2}$. Thus, we must also verify that $\sqrt{2x+3} - 1$ is not zero at any of the points where $\lfloor 4x^2-36 \rfloor = 0$.

$\sqrt{2x+3} - 1 = 0$ implies $\sqrt{2x+3} = 1$, so $2x+3 = 1$, which gives $2x = -2$, and $x = -1$. Since $-1$ is not in our exclusion interval $[3, \frac{\sqrt{37}}{2})$, this is not a problem.

Combining the conditions $x \ge -\frac{3}{2}$ and $x \notin [3, \frac{\sqrt{37}}{2})$, the domain of $g(x)$ is $[-\frac{3}{2}, \infty) \setminus [3, \frac{\sqrt{37}}{2})$, which is $[-\frac{3}{2}, 3) \cup [\frac{\sqrt{37}}{2}, \infty)$.

### 1.2 (b) Discussing the continuity of g(x) at a = -3/2

At $a = -\frac{3}{2}$, we have $g(-\frac{3}{2}) = \frac{\sqrt{2(-\frac{3}{2})+3}-1}{\lfloor 4(-\frac{3}{2})^2 - 36 \rfloor} = \frac{\sqrt{-3+3}-1}{\lfloor 4(\frac{9}{4})-36 \rfloor} = \frac{-1}{\lfloor 9-36 \rfloor} = \frac{-1}{\lfloor -27 \rfloor} = \frac{-1}{-27} = \frac{1}{27}$.

Now we examine the limit as $x$ approaches $-\frac{3}{2}$ from the right. Since $x \ge -\frac{3}{2}$ must be true, $x$ must approach $-\frac{3}{2}$ from the right.

$\lim_{x \to -\frac{3}{2}^+} g(x) = \lim_{x \to -\frac{3}{2}^+} \frac{\sqrt{2x+3} - 1}{\lfloor 4x^2 - 36 \rfloor} = \frac{\sqrt{2(-\frac{3}{2})+3}-1}{\lfloor 4(-\frac{3}{2})^2 - 36 \rfloor} = \frac{1}{27}$.

Since $g(-\frac{3}{2}) = \lim_{x \to -\frac{3}{2}^+} g(x) = \frac{1}{27}$, $g(x)$ is right-continuous at $x = -\frac{3}{2}$.

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3. Final Answer

1.1

(a) The domain of $f(x)$ is $(-\infty, -5] \cup (-3, -1) \cup [2, \infty)$.

(b) $f(x)$ is right-continuous at $x=2$.

1.2

(a) The domain of $g(x)$ is $[-\frac{3}{2}, 3) \cup [\frac{\sqrt{37}}{2}, \infty)$.

(b) $g(x)$ is right-continuous at $x = -\frac{3}{2}$.