The problem consists of 5 parts. 1.1. Given two functions $f(x)$ and $g(x)$, we need to find their domains and ranges. 1.2. Given a function $f(x)$, we need to find its domain and discuss the continuity of $f(x)$ at $x=0$. 1.3. Given a piecewise function $f(x)$, we need to discuss its continuity at $a=0$. 1.4. We need to discuss the continuity of $f(x) = \sqrt{x}$ at $a=0$. 1.5. We need to determine if the function $f(x) = \frac{2-|x|}{2-x}$ is left continuous at $a=-2$, and explain why or why not.

1. Problem Description

The problem consists of 5 parts.

1.

1. Given two functions $f(x)$ and $g(x)$, we need to find their domains and ranges.

1.

2. Given a function $f(x)$, we need to find its domain and discuss the continuity of $f(x)$ at $x=0$.

1.

3. Given a piecewise function $f(x)$, we need to discuss its continuity at $a=0$.

1.

4. We need to discuss the continuity of $f(x) = \sqrt{x}$ at $a=0$.

1.

5. We need to determine if the function $f(x) = \frac{2-|x|}{2-x}$ is left continuous at $a=-2$, and explain why or why not.

2. Solution Steps

1.1

a) Domain of $f(x)$:

If $x<0$, $f(x) = [[x]]+1$. This is defined for all $x<0$.

If $x>0$, $f(x) = x+1$. This is defined for all $x>0$.

If $x \ge 1$, $f(x) = ln(x-2)$. Since we must have $x-2>0$, so $x>2$. Combined with $x \ge 1$, we have $x>2$.

Therefore, $D_f = (-\infty, 0) \cup (0, \infty) \cup (2, \infty) = (-\infty, \infty)$.

b) Range of $f(x)$:

If $x<0$, $f(x) = [[x]]+1$. Since $[[x]]$ takes all integer values less than 0, $f(x)$ takes all integer values less than or equal to

0. If $x>0$, $f(x) = x+1$. Then $f(x) > 1$.

If $x>2$, $f(x) = ln(x-2)$. The range of $ln(x-2)$ when $x>2$ is $(-\infty, \infty)$.

Therefore, $R_f = \{ integers \le 0 \} \cup (1, \infty)$.

c) Domain of $g(x)$:

$g(x) = \frac{1}{x^2-4}$ if $x<1$.

We need $x^2-4 \ne 0$, so $x \ne \pm 2$.

Since $x<1$, the domain is $(-\infty, -2) \cup (-2, 1)$.

d) Range of $g(x)$:

Since $x<1$, $x^2 < 1$. Thus $x^2 - 4 < -3$.

Therefore, $\frac{1}{x^2-4} > -\frac{1}{3}$.

Let $y = \frac{1}{x^2-4}$. Then $x^2-4 = \frac{1}{y}$, so $x^2 = 4+\frac{1}{y}$.

Since $x^2 \ge 0$, we need $4+\frac{1}{y} \ge 0$, which means $\frac{1}{y} \ge -4$, or $y \le -\frac{1}{4}$ or $y>0$.

$g(x)$ is continuous in $(-\infty, -2)$ and $(-2, 1)$.

When $x$ approaches 1, $g(x) \rightarrow \frac{1}{1-4} = -\frac{1}{3}$.

When $x$ approaches -2 from left, $g(x) \rightarrow -\infty$. When $x$ approaches -2 from right, $g(x) \rightarrow \infty$.

Thus the range is $(-\infty, -\frac{1}{4}] \cup (0, \infty)$.

1.2

a) Domain of $f(x) = \frac{1}{[[3x-2]]}$:

We need $[[3x-2]] \ne 0$.

So $3x-2 \notin [0, 1)$.

$3x-2 < 0$ or $3x-2 \ge 1$.

$3x < 2$ or $3x \ge 3$.

$x < \frac{2}{3}$ or $x \ge 1$.

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

b) Continuity of $f(x)$ at $x=0$:

$3x-2 = -2$ at $x=0$, so $[[3x-2]] = [[-2]] = -2$.

Thus, $f(0) = \frac{1}{-2} = -\frac{1}{2}$.

Since $x=0 < \frac{2}{3}$, $f(x)$ is defined at $x=0$.

We can find the left-hand limit and right-hand limit to determine the continuity.

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

2. When $x< \frac{2}{3}$, $[[3x-2]] = -2$, $f(x) = -\frac{1}{2}$.

When $x \ge 1$, $[[3x-2]] = -2$, $f(x) = -\frac{1}{2}$.

1.3

Continuity of $f(x)$ at $a=0$:

If $-1 < x < 1$, $f(x) = 1-x^2$.

Left-hand limit at $x=0$:

Since $-1 < x < 1$ as $x$ approaches 0 from left, $f(x) = 1-x^2$.

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

Right-hand limit at $x=0$:

Since $-1 < x < 1$ as $x$ approaches 0 from right, $f(x) = 1-x^2$.

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

Since the function is continuous at $x=0$, $f(0) = 1 - 0^2 = 1$, $lim_{x \to 0} f(x) = 1$. Therefore $f(x)$ is continuous at $x=0$.

1.4

Continuity of $f(x) = \sqrt{x}$ at $a=0$:

For $x<0$, $f(x) = \sqrt{x}$ is not defined. The domain of $\sqrt{x}$ is $x \ge 0$.

$f(0) = \sqrt{0} = 0$.

$lim_{x \to 0^+} \sqrt{x} = 0$.

Since the left-hand limit is not defined and the right-hand limit exists and equals the function value at $x=0$, the function is right continuous at $x=0$.

1.5

Left continuity of $f(x) = \frac{2-|x|}{2-x}$ at $a=-2$:

$f(-2) = \frac{2-|-2|}{2-(-2)} = \frac{2-2}{4} = 0$.

For $x<0$, $|x| = -x$.

$f(x) = \frac{2+x}{2-x}$.

The left-hand limit at $x=-2$ is

$lim_{x \to -2^-} \frac{2+x}{2-x} = \frac{2-2}{2-(-2)} = \frac{0}{4} = 0$.

Since $lim_{x \to -2^-} f(x) = f(-2)$, the function is left continuous at $x=-2$.

3. Final Answer

1.1

a) $D_f = (-\infty, \infty)$

b) $R_f = \{ integers \le 0 \} \cup (1, \infty)$

c) $D_g = (-\infty, -2) \cup (-2, 1)$

d) $R_g = (-\infty, -\frac{1}{4}] \cup (0, \infty)$

1.2

a) $D_f = (-\infty, \frac{2}{3}) \cup [1, \infty)$

b) Continuous at x=0

1.3

Continuous at a=0

1.4

Right continuous at a=0

1.5

Yes, left continuous at a=-2