The problem consists of three parts: 1. Sketch the graph of the relation $R = A \times B$ for given sets $A$ and $B$. Determine if $R$ is a function from $A$ to $B$. If not, provide a subset of $R$ that is a function.

1. Problem Description

The problem consists of three parts:

1. Sketch the graph of the relation $R = A \times B$ for given sets $A$ and $B$. Determine if $R$ is a function from $A$ to $B$. If not, provide a subset of $R$ that is a function.

2. Determine if a given relation defines $f$ as a function of $x$. If it is a function, find its domain.

3. Find the domain of given functions.

2. Solution Steps

1. a) $A = Z$ and $B = R$. $R = A \times B = \{(a, b) \mid a \in Z, b \in R\}$.

This relation is not a function, since for any $a \in A$, there are infinitely many $b \in B$ such that $(a,b) \in R$. To create a subset that is a function, we can define $f(x) = 0$ for all $x \in Z$. Then $f = \{(x, 0) \mid x \in Z\}$ is a subset of $R$ that is a function.

b) $A = \{x \in R \mid |x| > 1\}$ and $B = [-1, 1]$. $R = A \times B = \{(a, b) \mid a \in R, |a| > 1, b \in [-1, 1]\}$.

This is not a function because for a given $a \in A$, there are infinitely many values $b \in [-1, 1]$ such that $(a, b) \in R$. A subset that is a function can be $f = \{(x, 0) \mid x \in A\}$.

c) $A = [-3, 1)$ and $B = \{x \in Z \mid 2x^2 - 5x - 3 = 0\}$. We solve $2x^2 - 5x - 3 = 0$.

$(2x + 1)(x - 3) = 0$. So $x = -\frac{1}{2}$ or $x = 3$. Thus, $B = \{-\frac{1}{2}, 3\}$.

Since $B \subset Z$, $B = \emptyset$. $R = A \times B = \emptyset$.

The empty set is a function because it satisfies the definition vacuously.

d) $A = \{1, 2, 3, 4\}$ and $B = \{-1, 0, 1\}$. $R = A \times B = \{(a, b) \mid a \in A, b \in B\} = \{(1, -1), (1, 0), (1, 1), (2, -1), (2, 0), (2, 1), (3, -1), (3, 0), (3, 1), (4, -1), (4, 0), (4, 1)\}$.

This is not a function, since for any $a \in A$, there are multiple values in B. A subset that is a function can be $f = \{(1, -1), (2, 0), (3, 1), (4, -1)\}$.

2. a)

$f(x) = \begin{cases} \sqrt{x^4 - 16} & \text{if } x \geq 0 \\ 6 - x & \text{if } x^2 \leq 4 \end{cases}$

We have two conditions on $x$: $x \geq 0$ and $x^2 \leq 4 \implies -2 \leq x \leq 2$.

If $x \geq 0$, then $x^4 - 16 \geq 0$, so $x^4 \geq 16$ and $x \geq 2$. Combining with $x \geq 0$, we have $x \geq 2$.

If $x^2 \leq 4$, then $-2 \leq x \leq 2$.

The domain of $f$ is the union of the intervals where the conditions are satisfied. We have $x \geq 2$ and $-2 \leq x \leq 2$. Therefore, the domain of $f$ is $\{-2 \le x \le 2\} \cup \{x \ge 2\} = [-2, -\infty)$.

However, since $x$ cannot be both $\geq 0$ and $x^2 \le 4$ at the same time (or they can at $x=2$), we must check if this function is well defined when $x=2$:

If $x = 2$, then $f(2) = \sqrt{2^4 - 16} = 0$ and $f(2) = 6 - 2 = 4$. Since we get two different values for $f(2)$, this is not a function.

b)

$f(x) = \begin{cases} \sqrt{x^4 - 16} & \text{if } x \geq 0 \\ 2 - x & \text{if } x^2 \leq 4 \end{cases}$

If $x \geq 0$, then $x^4 - 16 \geq 0$, so $x^4 \geq 16$ and $x \geq 2$. Combining with $x \geq 0$, we have $x \geq 2$.

If $x^2 \leq 4$, then $-2 \leq x \leq 2$.

The domain of $f$ is the union of the intervals where the conditions are satisfied. We have $x \geq 2$ and $-2 \leq x \leq 2$. Therefore, the domain of $f$ is $[-2, 2] \cup [2, \infty)$.

If $x = 2$, then $f(2) = \sqrt{2^4 - 16} = 0$ and $f(2) = 2 - 2 = 0$. Since we get the same value for $f(2)$, this could be a function.

We have $x \geq 2$ and $-2 \leq x \leq 2$, so the domain is $[-2, 2] \cup [2, \infty) = [-2, \infty)$. But, the condition $x \geq 2$ must hold for $\sqrt{x^4 - 16}$ to be real. Therefore, the domain is $[-2, 2] \cup [2, \infty) = [-2, 2] \cup [2, \infty)$, but $x\ge 2$ must be valid to use $\sqrt{x^4-16}$. Thus $x \in [-2,2]$ uses $2-x$ and $x\ge 2$ uses $\sqrt{x^4-16}$.

Then, for $x\ge 2$, $f(x) = \sqrt{x^4-16}$. For $x\in [-2, 2]$, $f(x) = 2-x$. $x = 2$ is a possible value for either of these. Then, this is a function with the domain $[-2, \infty)$.

c) $x^2 f(x) + x (f(x))^2 = 2$.

Let $y = f(x)$. Then $x^2 y + x y^2 = 2$, or $xy^2 + x^2 y - 2 = 0$. Solving for $y$, we use the quadratic formula:

$y = \frac{-x^2 \pm \sqrt{x^4 - 4x(-2)}}{2x} = \frac{-x^2 \pm \sqrt{x^4 + 8x}}{2x}$.

For this to be a function, we need $x^4 + 8x \geq 0$, $x(x^3 + 8) \geq 0$. So $x \geq 0$ or $x^3 \leq -8$, which means $x \leq -2$. If $x=0$, the equation is undefined, so $x \ne 0$.

Since there are two possible values for $y$ for each $x$ (where the function is defined), this is not a function.

d) $x^2 f(x) + x (f(x))^2 = 2$ and $f(x) + \frac{x}{2} < 0$.

We know that $f(x) = \frac{-x^2 \pm \sqrt{x^4 + 8x}}{2x}$. Also, $f(x) < -\frac{x}{2}$.

So, $\frac{-x^2 \pm \sqrt{x^4 + 8x}}{2x} < -\frac{x}{2}$.

Multiply by 2x. If $x > 0$, $-x^2 \pm \sqrt{x^4 + 8x} < -x^2$ which implies $\pm \sqrt{x^4 + 8x} < 0$. Then $-\sqrt{x^4 + 8x} < 0$, so we only consider $f(x) = \frac{-x^2 - \sqrt{x^4 + 8x}}{2x}$.

If $x < 0$, then $-x^2 \pm \sqrt{x^4 + 8x} > -x^2$ which implies $\pm \sqrt{x^4 + 8x} > 0$. Then $\sqrt{x^4 + 8x} > 0$, so we only consider $f(x) = \frac{-x^2 + \sqrt{x^4 + 8x}}{2x}$.

3. a) $f(x) = \sqrt{\frac{ln(x^2 - 1)}{x^x - x}}$

We need $\frac{ln(x^2 - 1)}{x^x - x} \geq 0$. Also, $x^2 - 1 > 0$, so $x^2 > 1$ or $|x| > 1$. $x^x - x \neq 0$.

Case 1: $ln(x^2 - 1) \geq 0$ and $x^x - x > 0$. $ln(x^2 - 1) \geq 0 \implies x^2 - 1 \geq 1 \implies x^2 \geq 2 \implies |x| \geq \sqrt{2}$.

Case 2: $ln(x^2 - 1) \leq 0$ and $x^x - x < 0$. $ln(x^2 - 1) \leq 0 \implies x^2 - 1 \leq 1 \implies x^2 \leq 2 \implies |x| \leq \sqrt{2}$.

Combining $|x| > 1$ and $|x| \geq \sqrt{2}$, we get $x \geq \sqrt{2}$ or $x \leq -\sqrt{2}$. $x^x - x > 0 \implies x^x > x$.

Combining $|x| > 1$ and $|x| \leq \sqrt{2}$, we get $1 < |x| \leq \sqrt{2}$. $x^x - x < 0 \implies x^x < x$.

b) $h(x) = \sqrt[6]{(\sqrt{2x})^x - (2x)^{\sqrt{x}}}$

We need $(\sqrt{2x})^x - (2x)^{\sqrt{x}} \geq 0$, which means $(\sqrt{2x})^x \geq (2x)^{\sqrt{x}}$. Also, $2x \geq 0$, so $x \geq 0$.

$(\sqrt{2x})^x \geq (2x)^{\sqrt{x}} \implies (2x)^{x/2} \geq (2x)^{\sqrt{x}}$. If $2x > 1$, $x/2 \geq \sqrt{x}$, so $x^2/4 \geq x$, $x^2 \geq 4x$, $x \geq 4$. If $0 < 2x < 1$, $x/2 \leq \sqrt{x}$, so $0 \leq x < 1/2$. Thus the domain is $[4, \infty) \cup [0, 1/2)$. However, we must exclude the case where $2x=1$, since then $1^{1/2} \ge 1^1$ is valid. $x = 1/2$ must be included. Thus the domain is $[4, \infty) \cup [0, 1/2]$. Also when $x = 0$, then the domain is valid, so the domain is $x = 0 \cup [4,\infty) \cup [0, 1/2]$. This gives $[0,1/2] \cup [4,\infty)$.

3. Final Answer

1. a) Not a function. $f = \{(x, 0) \mid x \in Z\}$

b) Not a function. $f = \{(x, 0) \mid x \in A\}$

c) Function. $R = \emptyset$

d) Not a function. $f = \{(1, -1), (2, 0), (3, 1), (4, -1)\}$

2. a) Not a function.

b) Domain: $[-2, \infty)$

3. a) $\{x \in R \mid (x \leq -\sqrt{2} \text{ or } x \geq \sqrt{2}) \text{ and } x^x > x\} \cup \{x \in R \mid (1 < |x| \leq \sqrt{2}) \text{ and } x^x < x\}$

b) $[0, \frac{1}{2}] \cup [4, \infty)$