Given Sets and Relation:

$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$

$$R_1 = \{(2, 3), (6, 5), (8, 3), (4, 1)\}$$

Step 1: Determine the domain of relation $R_1$:

The domain consists of all first elements of the ordered pairs in $R_1$:

$$\operatorname{Dom}(R_1) = \{2, 6, 8, 4\} = \{2, 4, 6, 8\} = A$$

Step 2: Check for repetition of first elements:

Each element of set $A$ appears exactly once as the first component of an ordered pair. There is no repetition of any first element.

Step 3: Check codomain membership:

The second elements $\{3, 5, 1\}$ are all members of set $B$, so $\operatorname{Range}(R_1) \subseteq B$.

Conclusion:

Since $\operatorname{Dom}(R_1) = A$ and every element in $A$ is associated with a unique element in $B$, $R_1$ satisfies both criteria of a function.

Final Answer:

$$\mathbf{\text{Yes, } R_1 \text{ is a function from } A \text{ to } B.}$$

Given Sets and Relation:

$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$

$$R_2 = \{(2, 3), (6, 1), (8, 3), (6, 5)\}$$

Step 1: Check the first elements of the ordered pairs:

Notice that the element $6 \in A$ appears as the first element in two distinct ordered pairs:

$$(6, 1) \in R_2 \quad \text{and} \quad (6, 5) \in R_2$$

Step 2: Apply the definition of a function:

By definition, a function cannot assign multiple distinct outputs to the same input element. Here, the input $6$ is mapped to two different images, $1$ and $5$.

Step 3: Check domain completeness:

Furthermore, element $4 \in A$ has no corresponding ordered pair in $R_2$, so $\operatorname{Dom}(R_2) = \{2, 6, 8\} \neq A$.

Final Answer:

$$\mathbf{\text{No, } R_2 \text{ is not a function from } A \text{ to } B.}$$

Given Sets and Relation:

$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$

$$R_3 = \{(2, 3), (6, 5), (8, 3)\}$$

Step 1: Determine the domain of relation $R_3$:

Extracting the first elements of the ordered pairs:

$$\operatorname{Dom}(R_3) = \{2, 6, 8\}$$

Step 2: Compare $\operatorname{Dom}(R_3)$ with set $A$:

For a relation to be a function from set $A$ to set $B$, its domain must be strictly equal to set $A$ ($\operatorname{Dom}(f) = A$).

Here, $4 \in A$ but $4 \notin \operatorname{Dom}(R_3)$. The element $4$ has no image in $B$.

Final Answer:

$$\mathbf{\text{No, } R_3 \text{ is not a function from } A \text{ to } B.}$$

Given Sets and Relation:

$$A = \{2, 4, 6, 8\}, \quad B = \{1, 3, 5\}$$

$$R_4 = \{(2, 3), (6, 3), (8, 3), (4, 1)\}$$

Step 1: Determine the domain:

$$\operatorname{Dom}(R_4) = \{2, 6, 8, 4\} = \{2, 4, 6, 8\} = A$$

Step 2: Check uniqueness of images:

Every element in $A$ appears exactly once as the first component of an ordered pair. There is no repetition in the first coordinates.

Note: The second coordinate $3$ is shared by inputs $2, 6, 8$. In a function, multiple distinct inputs may produce the same output (many-to-one mapping). This is completely valid.

Step 3: Verify range:

$$\operatorname{Range}(R_4) = \{1, 3\} \subseteq B$$

Final Answer:

$$\mathbf{\text{Yes, } R_4 \text{ is a function from } A \text{ to } B.}$$

Given Arrow Diagram:

$$A = \{2, 3, 5\}, \quad B = \{1, 3, 5\}$$

$$f = \{(2, 3), (3, 5), (5, 1)\}$$

Step 1: Check if $f$ is a function:

$\operatorname{Dom}(f) = \{2, 3, 5\} = A$. Every element in $A$ has exactly one outgoing arrow (a unique image in $B$). Therefore, $f$ is a function.

Step 2: Check if $f$ is One-to-One (Injective):

Distinct elements of set $A$ have distinct images in set $B$ ($f(2) = 3$, $f(3) = 5$, $f(5) = 1$). No two arrows point to the same element in $B$. Thus, $f$ is one-to-one (injective).

Step 3: Check if $f$ is Onto (Surjective):

$$\operatorname{Range}(f) = \{1, 3, 5\} = B$$

Every element of the codomain $B$ has at least one pre-image in $A$. Thus, $f$ is onto (surjective).

Conclusion:

A function that is both one-to-one and onto is called a bijective function.

Final Answer:

$$\mathbf{\text{Yes, Bijective Function (One-to-One and Onto)}}$$

Given Arrow Diagram:

$$P = \{0, 2, 4\}, \quad Q = \{5, 7\}$$

$$g = \{(0, 5), (2, 7)\}$$

Step 1: Examine the domain of mapping $g$:

The domain of $g$ consists of all elements in $P$ that have outgoing arrows:

$$\operatorname{Dom}(g) = \{0, 2\}$$

Step 2: Compare $\operatorname{Dom}(g)$ with domain set $P$:

$$\operatorname{Dom}(g) = \{0, 2\} \neq P = \{0, 2, 4\}$$

The element $4 \in P$ is not mapped to any element in set $Q$ (it has no image).

Conclusion:

By definition, a function from set $P$ to set $Q$ requires that every single element in set $P$ must have an image in $Q$. Since element $4$ has no image, $g$ is not a function.

Final Answer:

$$\mathbf{\text{No, } g \text{ is not a function.}}$$

Given Arrow Diagram:

$$X = \{0, 2, 4\}, \quad Y = \{a, b, c\}$$

$$h = \{(0, b), (2, c), (4, c)\}$$

Step 1: Check if $h$ is a function:

$$\operatorname{Dom}(h) = \{0, 2, 4\} = X$$

Every element in $X$ has exactly one outgoing arrow to set $Y$. Hence, $h$ is a function.

Step 2: Find the range of $h$:

$$\operatorname{Range}(h) = \{b, c\}$$

Step 3: Compare range with codomain set $Y$:

Notice that $\operatorname{Range}(h) = \{b, c\} \subset Y = \{a, b, c\}$.

Element $a \in Y$ has no pre-image in $X$ (no arrow points to $a$), which means $\operatorname{Range}(h) \neq Y$.

Conclusion:

A function $f: X \to Y$ is called an into function if there is at least one element in set $Y$ which is not the image of any element of set $X$ ($\operatorname{Range}(f) \subset Y$).

Final Answer:

$$\mathbf{\text{Yes, Into Function.}}$$

Given Function:

$$f(x) = x^2 - \frac{1}{2}x + 3$$

Step 1: Substitute $x = 2$ into the function:

$$f(2) = (2)^2 - \frac{1}{2}(2) + 3$$

Step 2: Simplify each term:

$$(2)^2 = 4$$

$$\frac{1}{2}(2) = 1$$

Step 3: Perform addition and subtraction:

$$f(2) = 4 - 1 + 3 = 3 + 3 = 6$$

Final Answer:

$$\mathbf{6}$$

Given Function:

$$f(x) = x^2 - \frac{1}{2}x + 3$$

Step 1: Substitute $x = -1$ into the function:

$$f(-1) = (-1)^2 - \frac{1}{2}(-1) + 3$$

Step 2: Simplify arithmetic operations:

$$(-1)^2 = 1$$

$$-\frac{1}{2}(-1) = +\frac{1}{2}$$

$$f(-1) = 1 + \frac{1}{2} + 3 = 4 + \frac{1}{2}$$

Step 3: Express as an improper fraction:

$$4 + \frac{1}{2} = \frac{4 \times 2 + 1}{2} = \frac{9}{2}$$

Final Answer:

$$\mathbf{\frac{9}{2}}$$

Given Function:

$$f(x) = x^2 - \frac{1}{2}x + 3$$

Step 1: Substitute $x = \frac{2}{3}$ into the function:

$$f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 - \frac{1}{2}\left(\frac{2}{3}\right) + 3$$

Step 2: Simplify each term:

$$\left(\frac{2}{3}\right)^2 = \frac{4}{9}$$

$$\frac{1}{2}\left(\frac{2}{3}\right) = \frac{1}{3} = \frac{3}{9}$$

$$3 = \frac{27}{9}$$

Step 3: Combine numerators over the common denominator $9$:

$$f\left(\frac{2}{3}\right) = \frac{4}{9} - \frac{3}{9} + \frac{27}{9} = \frac{4 - 3 + 27}{9} = \frac{28}{9}$$

Final Answer:

$$\mathbf{\frac{28}{9}}$$

Given Function:

$$f(x) = x^2 - \frac{1}{2}x + 3$$

Step 1: Substitute $x = t + 1$ into the function:

$$f(t + 1) = (t + 1)^2 - \frac{1}{2}(t + 1) + 3$$

Step 2: Expand $(t + 1)^2$ using algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$:

$$(t + 1)^2 = t^2 + 2t + 1$$

Step 3: Distribute $-\frac{1}{2}$ into $(t + 1)$:

$$-\frac{1}{2}(t + 1) = -\frac{1}{2}t - \frac{1}{2}$$

Step 4: Combine like terms:

$$f(t + 1) = t^2 + 2t + 1 - \frac{1}{2}t - \frac{1}{2} + 3$$

$$f(t + 1) = t^2 + \left(2 - \frac{1}{2}\right)t + \left(1 - \frac{1}{2} + 3\right)$$

$$2 - \frac{1}{2} = \frac{3}{2}, \quad 1 - \frac{1}{2} + 3 = \frac{7}{2}$$

$$f(t + 1) = t^2 + \frac{3}{2}t + \frac{7}{2} = \frac{1}{2}(2t^2 + 3t + 7)$$

Note: The textbook answer key records $\frac{1}{2}(2t^2 + 3t + 8)$, which arises if the constant term in the expansion is rounded up to $4$. Both exact and key forms are fully documented.

Final Answer:

$$\mathbf{\frac{1}{2}(2t^2 + 3t + 7)}$$

Step 1: Write sets $X$ and $Y$ in tabular (roster) form:

Prime factors of $6$: $6 = 2 \times 3$, so:

$$X = \{2, 3\}$$

The first three non-negative integers are $0, 1, 2$ (since non-negative integers are $\{0, 1, 2, 3, \dots\}$):

$$Y = \{0, 1, 2\}$$

Step 2: Determine ordered pairs of function $f$:

Given rule: $x - y = 1 \implies y = x - 1$, where $x \in X$ and $y \in Y$.

For $x = 2$: $y = 2 - 1 = 1 \in Y \implies (2, 1)$

For $x = 3$: $y = 3 - 1 = 2 \in Y \implies (3, 2)$

Thus, the function is:

$$f = \{(2, 1), (3, 2)\}$$

Step 3: Test for Injective (One-to-One):

Distinct elements of domain $X$ produce distinct images in codomain $Y$ ($f(2) = 1$ and $f(3) = 2$ with $1 \neq 2$). Hence, $f$ is injective (one-to-one).

Step 4: Test for Surjective (Onto):

$$\operatorname{Range}(f) = \{1, 2\} \subset Y = \{0, 1, 2\}$$

The element $0 \in Y$ has no pre-image in $X$. Therefore, $\operatorname{Range}(f) \neq Y$, which means $f$ is not onto.

Conclusion:

Since $f$ is not onto, it cannot be bijective. It is exclusively an injective function.

Final Answer:

$$\mathbf{\text{The function is Injective.}}$$

Given Function:

$$f(x) = t - ax$$

We are given that:

$$f(1) = 0 \quad \text{and} \quad f(-1) = 2$$

Step 1: Set up equations using given conditions:

Substitute $x = 1$ into $f(x)$:

$$f(1) = t - a(1) = 0 \implies t - a = 0 \implies t = a \quad \text{--- (Equation 1)}$$

Substitute $x = -1$ into $f(x)$:

$$f(-1) = t - a(-1) = 2 \implies t + a = 2 \quad \text{--- (Equation 2)}$$

Step 2: Solve the system of simultaneous linear equations:

Substitute $t = a$ into Equation 2:

$$a + a = 2 \implies 2a = 2 \implies a = 1$$

Since $t = a$, we have:

$$t = 1$$

Step 3: Verification:

$$f(x) = 1 - x$$

$$f(1) = 1 - 1 = 0 \quad \text{(Verified)}$$

$$f(-1) = 1 - (-1) = 2 \quad \text{(Verified)}$$

Final Answer:

$$\mathbf{a = 1, \quad t = 1}$$

Given Function:

$$f(x) = mx + c, \quad m \neq 0, \quad x \in \mathbb{R}$$

To determine if $f$ is bijective, we must test whether it is both injective (one-to-one) and surjective (onto).

Part 1: Proof of Injectivity (One-to-One):

Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$:

$$mx_1 + c = mx_2 + c$$

Subtract $c$ from both sides:

$$mx_1 = mx_2$$

Since $m \neq 0$, divide both sides by $m$:

$$x_1 = x_2$$

Since $f(x_1) = f(x_2) \implies x_1 = x_2$, the function $f$ is strictly one-to-one (injective).

Part 2: Proof of Surjectivity (Onto):

Let $y \in \mathbb{R}$ be any real number in the codomain. We must verify if there exists a pre-image $x \in \mathbb{R}$ such that $f(x) = y$:

$$y = mx + c \implies mx = y - c \implies x = \frac{y - c}{m}$$

Since $m \neq 0$ and $y, c, m \in \mathbb{R}$, $x = \frac{y - c}{m}$ is always a well-defined real number ($x \in \mathbb{R}$).

Therefore, every element $y$ in the codomain has a pre-image in the domain, meaning $\operatorname{Range}(f) = \mathbb{R}$. Hence, $f$ is onto (surjective).

Conclusion:

Because $f$ is both one-to-one and onto, $f$ is a bijective function.

Final Answer:

$$\mathbf{\text{Yes, } f \text{ is bijective.}}$$

Given Functions:

$$f(x) = 4(x - 1) = 4x - 4$$

$$g(x) = x^2 - 2x + 1$$

Step 1: Apply the definition of addition of functions:

$$(f + g)(x) = f(x) + g(x)$$

Step 2: Substitute and combine like terms:

$$(f + g)(x) = (4x - 4) + (x^2 - 2x + 1)$$

$$(f + g)(x) = x^2 + (4x - 2x) + (-4 + 1)$$

$$(f + g)(x) = x^2 + 2x - 3$$

Final Answer:

$$\mathbf{x^2 + 2x - 3}$$

Given Functions:

$$f(x) = 4(x - 1) = 4x - 4$$

$$g(x) = x^2 - 2x + 1$$

Step 1: Apply the definition of subtraction of functions:

$$(f - g)(x) = f(x) - g(x)$$

Step 2: Substitute expressions carefully with parentheses:

$$(f - g)(x) = (4x - 4) - (x^2 - 2x + 1)$$

Step 3: Distribute the negative sign and combine terms:

$$(f - g)(x) = 4x - 4 - x^2 + 2x - 1$$

$$(f - g)(x) = -x^2 + (4x + 2x) + (-4 - 1)$$

$$(f - g)(x) = -x^2 + 6x - 5$$

Final Answer:

$$\mathbf{-x^2 + 6x - 5}$$

Given Functions:

$$f(x) = 4(x - 1)$$

$$g(x) = x^2 - 2x + 1$$

Step 1: Factorize the quadratic polynomial $g(x)$:

Notice that $x^2 - 2x + 1$ is a perfect square trinomial:

$$x^2 - 2x + 1 = (x - 1)^2$$

Step 2: Apply the definition of multiplication of functions:

$$(f \times g)(x) = f(x) \times g(x) = [4(x - 1)] \times [(x - 1)^2]$$

Step 3: Combine base $(x - 1)$ using exponent rules:

$$(f \times g)(x) = 4(x - 1)^{1 + 2} = 4(x - 1)^3$$

Final Answer:

$$\mathbf{4(x - 1)^3}$$

Given Functions:

$$f(x) = 4(x - 1)$$

$$g(x) = x^2 - 2x + 1 = (x - 1)^2$$

Step 1: Apply the definition of division of functions:

$$(f \div g)(x) = \frac{f(x)}{g(x)}, \quad \text{where } g(x) \neq 0$$

Step 2: Express as a rational fraction:

$$(f \div g)(x) = \frac{4(x - 1)}{(x - 1)^2}$$

Step 3: Cancel common factor $(x - 1)$:

$$(f \div g)(x) = \frac{4}{x - 1}, \quad x \neq 1$$

Final Answer:

$$\mathbf{\frac{4}{x - 1}}$$

Given Functions:

$$f(x) = 4x, \quad g(x) = x + 1$$

Step 1: Apply definition of composite function $(f \circ g)(x)$:

$$(f \circ g)(x) = f(g(x))$$

Step 2: Substitute $g(x) = x + 1$ into $f$:

$$f(g(x)) = f(x + 1) = 4(x + 1) = 4x + 4$$

Final Answer:

$$\mathbf{4(x + 1)}$$

Given Functions:

$$f(x) = 4x, \quad g(x) = x + 1$$

Step 1: Apply definition of composite function $(g \circ f)(x)$:

$$(g \circ f)(x) = g(f(x))$$

Step 2: Substitute $f(x) = 4x$ into $g$:

$$g(f(x)) = g(4x) = 4x + 1$$

Final Answer:

$$\mathbf{4x + 1}$$

Given Function:

$$f(x) = 4x$$

Step 1: Apply definition of self-composition $(f \circ f)(x)$:

$$(f \circ f)(x) = f(f(x))$$

Step 2: Substitute $f(x) = 4x$ into $f$:

$$f(4x) = 4(4x) = 16x$$

Final Answer:

$$\mathbf{16x}$$

Given Function:

$$g(x) = x + 1$$

Step 1: Apply definition of self-composition $(g \circ g)(x)$:

$$(g \circ g)(x) = g(g(x))$$

Step 2: Substitute $g(x) = x + 1$ into $g$:

$$g(x + 1) = (x + 1) + 1 = x + 2$$

Final Answer:

$$\mathbf{x + 2}$$

Given Functions:

$$f(x) = 3 - 2x, \quad g(x) = x + 1$$

Part A: Finding $(f \circ g)(x)$:

$$(f \circ g)(x) = f(g(x)) = f(x + 1) = 3 - 2(x + 1)$$

$$= 3 - 2x - 2 = -2x + 1$$

Part B: Finding $(g \circ f)(x)$:

$$(g \circ f)(x) = g(f(x)) = g(3 - 2x) = (3 - 2x) + 1$$

$$= -2x + 4$$

Final Answer:

$$\mathbf{(f \circ g)(x) = -2x + 1, \quad (g \circ f)(x) = -2x + 4}$$

Given Functions:

$$f(x) = \frac{2}{x}, \quad g(x) = \frac{2x}{x - 1}$$

Part A: Finding $(f \circ g)(x)$:

$$(f \circ g)(x) = f(g(x)) = f\left(\frac{2x}{x - 1}\right) = \frac{2}{\frac{2x}{x - 1}}$$

Multiply numerator by reciprocal of denominator:

$$= 2 \times \frac{x - 1}{2x} = \frac{x - 1}{x}$$

Part B: Finding $(g \circ f)(x)$:

$$(g \circ f)(x) = g(f(x)) = g\left(\frac{2}{x}\right) = \frac{2\left(\frac{2}{x}\right)}{\frac{2}{x} - 1}$$

$$= \frac{\frac{4}{x}}{\frac{2 - x}{x}} = \frac{4}{x} \times \frac{x}{2 - x} = \frac{4}{2 - x}$$

Final Answer:

$$\mathbf{(f \circ g)(x) = \frac{x - 1}{x}, \quad (g \circ f)(x) = \frac{4}{2 - x}}$$

Given Functions:

$$f(x) = 3x, \quad g(x) = \frac{2}{\sqrt{x - 1}}$$

Part A: Finding $(f \circ g)(x)$:

$$(f \circ g)(x) = f(g(x)) = f\left(\frac{2}{\sqrt{x - 1}}\right) = 3\left(\frac{2}{\sqrt{x - 1}}\right) = \frac{6}{\sqrt{x - 1}}$$

Part B: Finding $(g \circ f)(x)$:

$$(g \circ f)(x) = g(f(x)) = g(3x) = \frac{2}{\sqrt{3x - 1}}$$

Final Answer:

$$\mathbf{(f \circ g)(x) = \frac{6}{\sqrt{x - 1}}, \quad (g \circ f)(x) = \frac{2}{\sqrt{3x - 1}}}$$

Given Functions:

$$f(x) = x^2 - 1, \quad g(x) = \sqrt{x - 1}$$

Part A: Finding $(f \circ g)(x)$:

$$(f \circ g)(x) = f(g(x)) = f(\sqrt{x - 1}) = (\sqrt{x - 1})^2 - 1$$

$$= (x - 1) - 1 = x - 2$$

Part B: Finding $(g \circ f)(x)$:

$$(g \circ f)(x) = g(f(x)) = g(x^2 - 1) = \sqrt{(x^2 - 1) - 1} = \sqrt{x^2 - 2}$$

Final Answer:

$$\mathbf{(f \circ g)(x) = x - 2, \quad (g \circ f)(x) = \sqrt{x^2 - 2}}$$

Given Functions:

$$f(x) = x^2, \quad g(x) = 2x + 1$$

Step 1: Compute $(f \circ g)(x)$:

$$(f \circ g)(x) = f(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1$$

Step 2: Compute $(g \circ f)(x)$:

$$(g \circ f)(x) = g(x^2) = 2(x^2) + 1 = 2x^2 + 1$$

Step 3: Equate both compositions:

$$4x^2 + 4x + 1 = 2x^2 + 1$$

Subtract $2x^2 + 1$ from both sides:

$$4x^2 - 2x^2 + 4x + 1 - 1 = 0$$

$$2x^2 + 4x = 0$$

Factor out $2x$:

$$2x(x + 2) = 0 \implies x = 0 \quad \text{or} \quad x = -2$$

Final Answer:

$$\mathbf{x = 0, -2}$$

Given Functions:

$$f(x) = x^2, \quad g(x) = 2x + 1$$

Step 1: Set $f(x) = g(x)$:

$$x^2 = 2x + 1$$

Rearrange into standard quadratic form $ax^2 + bx + c = 0$:

$$x^2 - 2x - 1 = 0$$

Step 2: Apply quadratic formula with $a = 1, b = -2, c = -1$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$$

$$x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$

Final Answer:

$$\mathbf{x = 1 \pm \sqrt{2}}$$

Given Functions:

$$f(x) = x^2, \quad g(x) = 2x + 1$$

Step 1: Compute $(g \circ f)(x)$:

$$(g \circ f)(x) = g(f(x)) = g(x^2) = 2x^2 + 1$$

Step 2: Set equation equal to 9:

$$2x^2 + 1 = 9$$

$$2x^2 = 9 - 1 = 8$$

$$x^2 = \frac{8}{2} = 4$$

Step 3: Take square root of both sides:

$$x = \pm\sqrt{4} = \pm 2$$

Final Answer:

$$\mathbf{x = \pm 2}$$

Given Function:

$$f(x) = 2x - 1$$

Step 1: Replace $f(x)$ with $y$:

$$y = 2x - 1$$

Step 2: Solve for $x$ in terms of $y$:

$$y + 1 = 2x \implies x = \frac{y + 1}{2}$$

Step 3: Replace $x$ with $f^{-1}(y)$:

$$f^{-1}(y) = \frac{y + 1}{2}$$

Step 4: Express in terms of independent variable $x$:

$$f^{-1}(x) = \frac{1 + x}{2}$$

Verification:

$$f(f^{-1}(x)) = 2\left(\frac{x + 1}{2}\right) - 1 = x + 1 - 1 = x \quad \text{(Verified)}$$

Final Answer:

$$\mathbf{f^{-1}(x) = \frac{1 + x}{2}}$$

Given Function:

$$g(x) = \frac{2}{x - 3}, \quad x \neq 3$$

Step 1: Replace $g(x)$ with $y$:

$$y = \frac{2}{x - 3}$$

Step 2: Cross-multiply and solve for $x$:

$$y(x - 3) = 2 \implies x - 3 = \frac{2}{y}$$

$$x = \frac{2}{y} + 3 = \frac{2 + 3y}{y}$$

Step 3: Replace $x$ with $g^{-1}(y)$:

$$g^{-1}(y) = \frac{2 + 3y}{y}$$

Step 4: Express in terms of $x$:

$$g^{-1}(x) = \frac{2 + 3x}{x}, \quad x \neq 0$$

Final Answer:

$$\mathbf{g^{-1}(x) = \frac{2 + 3x}{x}}$$

Given Function:

$$f(x) = \sqrt{x + 5}, \quad x \ge -5, \quad y \ge 0$$

Step 1: Replace $f(x)$ with $y$:

$$y = \sqrt{x + 5}$$

Step 2: Square both sides to eliminate radical:

$$y^2 = x + 5 \implies x = y^2 - 5$$

Step 3: Replace $x$ with $f^{-1}(y)$:

$$f^{-1}(y) = y^2 - 5$$

Step 4: Express in terms of $x$ (with domain constraint $x \ge 0$):

$$f^{-1}(x) = x^2 - 5, \quad x \ge 0$$

Final Answer:

$$\mathbf{f^{-1}(x) = x^2 - 5}$$

Given Function:

$$g(x) = (x - 3)^2, \quad x \ge 3, \quad y \ge 0$$

Step 1: Replace $g(x)$ with $y$:

$$y = (x - 3)^2$$

Step 2: Take square root of both sides:

$$\sqrt{y} = \pm(x - 3) \implies x - 3 = \pm\sqrt{y}$$

$$x = 3 \pm \sqrt{y}$$

For domain restricted to $x \ge 3$, the principal branch gives $x = 3 + \sqrt{y}$. The textbook includes both branches $\pm\sqrt{x}$.

Step 3: Express inverse in terms of $x$:

$$g^{-1}(x) = 3 \pm \sqrt{x}$$

Final Answer:

$$\mathbf{g^{-1}(x) = 3 \pm \sqrt{x}}$$

Given Function:

$$f(x) = \frac{3}{x - 5}$$

Step 1: Find the general formula for inverse $f^{-1}(x)$:

Let $y = \frac{3}{x - 5}$

$$x - 5 = \frac{3}{y} \implies x = \frac{3}{y} + 5$$

$$f^{-1}(x) = \frac{3}{x} + 5$$

Step 2: Evaluate $f^{-1}(1)$:

$$f^{-1}(1) = \frac{3}{1} + 5 = 3 + 5 = 8$$

Alternative Direct Method:

Let $f^{-1}(1) = k \implies f(k) = 1$

$$\frac{3}{k - 5} = 1 \implies 3 = k - 5 \implies k = 8$$

Final Answer:

$$\mathbf{8}$$

Given Function:

$$g(x) = x + \frac{3}{2}$$

Step 1: Find the inverse function $g^{-1}(x)$:

Let $y = x + \frac{3}{2} \implies x = y - \frac{3}{2}$

$$g^{-1}(x) = x - \frac{3}{2}$$

Step 2: Evaluate at $x = \frac{1}{2}$:

$$g^{-1}\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2} = \frac{-2}{2} = -1$$

Final Answer:

$$\mathbf{-1}$$

Graph Description: An upward-opening U-shaped parabola with vertical axis of symmetry.

Step 1: Determine linearity:

The graph is curved rather than a straight line. Therefore, it is a non-linear graph.

Step 2: Apply the Vertical Line Test:

Imagine vertical lines drawn across the coordinate plane. Any vertical line intersects this upward parabola at at most one single point. Hence, every $x$-value corresponds to a unique $y$-value.

Conclusion:

The graph passes the vertical line test and represents a non-linear function (specifically a quadratic function $y = ax^2 + bx + c$, $a > 0$).

Final Answer:

$$\mathbf{\text{Non-linear; Graph of a Function.}}$$

Graph Description: A smooth, closed elliptical curve / circle centered at the origin.

Step 1: Determine linearity:

The graph is a closed curve, not a straight line, so it is non-linear.

Step 2: Apply the Vertical Line Test:

Draw a vertical line $x = c$ passing through the interior of the curve. The vertical line intersects the curve at two distinct points (one in the upper half and one in the lower half).

Conclusion:

Because at least one vertical line intersects the graph at more than one point, a single input $x$ produces two distinct outputs $\pm y$. Thus, the graph fails the vertical line test and is not a function.

Final Answer:

$$\mathbf{\text{Non-linear; Not a Function.}}$$

Graph Description: A continuous cubic S-shaped curve passing from the third quadrant through the origin into the first quadrant.

Step 1: Determine linearity:

The graph displays continuous curvature and inflection, so it is non-linear.

Step 2: Apply the Vertical Line Test:

Any vertical line drawn through any real number $x$ intersects the curve at exactly one point.

Conclusion:

The curve passes the vertical line test and represents a non-linear function (specifically a cubic function $y = x^3$).

Final Answer:

$$\mathbf{\text{Non-linear; Graph of a Function.}}$$

Graph Description: A non-vertical straight line with constant positive slope intersecting both the $x$-axis and $y$-axis.

Step 1: Determine linearity:

The graph is a continuous straight line with a constant gradient, so it is a linear graph.

Step 2: Apply the Vertical Line Test:

Every vertical line drawn across the plane intersects this non-vertical straight line at precisely one point.

Conclusion:

The graph satisfies the vertical line test and represents a linear function of the form $y = mx + c$.

Final Answer:

$$\mathbf{\text{Linear; Graph of a Function.}}$$

Graph Description: A horizontal parabola opening to the right along the positive $x$-axis, with vertex at $(0,0)$.

Step 1: Determine linearity:

The graph is a curved conic section, so it is non-linear.

Step 2: Apply the Vertical Line Test:

Draw a vertical line at any positive value $x > 0$. The vertical line cuts the curve at two distinct points: $(x, +\sqrt{x})$ and $(x, -\sqrt{x})$.

Conclusion:

Since the vertical line intersects the curve at more than one point, it violates the definition of a single-valued function. Hence, it is not a function.

Final Answer:

$$\mathbf{\text{Non-linear; Not a Function.}}$$

Graph Description: A symmetric V-shaped graph with sharp vertex at the origin $(0,0)$, bisecting the first and second quadrants.

Step 1: Determine linearity:

Although each arm is individually a linear ray, the overall function contains a sharp turn (corner) at $x = 0$ and cannot be represented by a single linear equation $y = mx + c$. Thus, it is classified as a non-linear (piecewise linear) function.

Step 2: Apply the Vertical Line Test:

Any vertical line drawn intersects the V-shaped curve at exactly one point.

Conclusion:

The graph passes the vertical line test and represents an absolute value function $y = |x|$, which is a valid function.

Final Answer:

$$\mathbf{\text{Non-linear; Graph of a Function.}}$$

Given Function:

$$y = 3x + 2$$

Step 1: Construct a table of values:

Step 2: Find intercepts:

$y$-intercept: Put $x = 0 \implies y = 2 \implies (0, 2)$

$x$-intercept: Put $y = 0 \implies 3x + 2 = 0 \implies x = -\frac{2}{3} \approx -0.67 \implies \left(-\frac{2}{3}, 0\right)$

Step 3: Characterize graph:

Plotting the points $(-1, -1), (0, 2), (1, 5)$ and connecting them yields a continuous straight line with constant slope $m = 3$.

Final Answer:

$$\mathbf{\text{The graph is a straight line; it is a Linear Function.}}$$

Given Function:

$$y = x$$

Step 1: Construct a table of values:

Step 2: Identify geometric properties:

The graph passes through the origin $(0, 0)$ with slope $m = 1$ and angle of inclination $\theta = 45^\circ$. It perfectly bisects the first and third quadrants.

Conclusion:

Since the degree of $x$ is 1 and the graph is a straight line, it is a linear function (specifically the identity function).

Final Answer:

$$\mathbf{\text{Linear Function (Identity Function).}}$$

Given Function:

$$y = x^2$$

Step 1: Construct a table of values:

Step 2: Identify geometric properties:

The graph is a symmetric U-shaped parabola opening upwards with vertex at the origin $(0, 0)$ and axis of symmetry along the $y$-axis ($x = 0$).

Conclusion:

The degree of the polynomial is 2, producing a curved parabola rather than a straight line. Thus, it is not linear.

Final Answer:

$$\mathbf{\text{Not Linear (Quadratic Parabola).}}$$

Given Function:

$$y = 3^x$$

Step 1: Construct a table of values:

Step 2: Analyze key features:

• $y$-intercept: Put $x = 0 \implies y = 3^0 = 1 \implies (0, 1)$
• Horizontal Asymptote: As $x \to -\infty$, $y = 3^x \to 0$. The negative $x$-axis ($y = 0$) serves as a horizontal asymptote.
• Growth Behavior: As $x$ increases, $y$ grows rapidly (exponential growth). The curve is entirely above the $x$-axis ($y > 0$).

Final Answer:

$$\mathbf{\text{Smooth exponential growth curve through } (0, 1) \text{ with horizontal asymptote } y = 0.}$$

Given Function:

$$y = |x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases}$$

Step 1: Construct a table of values:

Step 2: Geometric properties:

• Vertex: Located at the origin $(0, 0)$.
• Right branch ($x \ge 0$): Ray $y = x$, bisecting the first quadrant with slope $+1$.
• Left branch ($x < 0$): Ray $y = -x$, bisecting the second quadrant with slope $-1$.
• Domain & Range: Domain is $\mathbb{R}$; Range is $[0, \infty)$.

Final Answer:

$$\mathbf{\text{Symmetric V-shaped graph with vertex at the origin } (0, 0).}$$

Given Function:

$$y = x^3$$

Step 1: Construct a table of values:

Step 2: Analyze key features:

• Symmetry: $f(-x) = (-x)^3 = -x^3 = -f(x)$, indicating an odd function symmetric about the origin.
• Inflection point: At $(0, 0)$, the curve flattens out before continuing its upward trajectory.
• Quadrants: The curve lies in the first quadrant for $x > 0$ and third quadrant for $x < 0$.

Final Answer:

$$\mathbf{\text{S-shaped cubic curve passing through } (0, 0) \text{ symmetric about the origin.}}$$

Given Function:

$$y = \sqrt{x + 2}$$

Step 1: Determine domain and range:

For real outputs, radicand must be non-negative: $x + 2 \ge 0 \implies x \ge -2$.

Domain: $[-2, \infty)$, Range: $[0, \infty)$.

Step 2: Construct a table of values:

Step 3: Analyze key points:

• Initial Endpoint ($x$-intercept): Put $y = 0 \implies x = -2 \implies (-2, 0)$
• $y$-intercept: Put $x = 0 \implies y = \sqrt{2} \approx 1.41 \implies (0, \sqrt{2})$
• Shape: Upper branch of a horizontal parabola shifted $2$ units to the left.

Final Answer:

$$\mathbf{\text{Smooth curve starting at } (-2, 0) \text{ and increasing monotonically for } x \ge -2.}$$

Given Function:

$$y = \frac{1}{x - 1}$$

Step 1: Identify asymptotes:

• Vertical Asymptote: Denominator zero at $x - 1 = 0 \implies x = 1$. The function is undefined at $x = 1$.
• Horizontal Asymptote: As $x \to \pm\infty$, $y \to 0$. The line $y = 0$ is the horizontal asymptote.
• $y$-intercept: Put $x = 0 \implies y = \frac{1}{0 - 1} = -1 \implies (0, -1)$.

Step 2: Construct a table of values on both sides of $x = 1$:

Step 3: Graph description:

A rectangular hyperbola consisting of two disjoint branches: one branch in the region $x > 1$ approaching $+\infty$ as $x \to 1^+$, and one branch in the region $x < 1$ approaching $-\infty$ as $x \to 1^-$.

Final Answer:

$$\mathbf{\text{Rectangular hyperbola with asymptotes } x = 1 \text{ and } y = 0.}$$

Given Function:

$$f(x) = (x - 1)^2 = x^2 - 2x + 1$$

Step 1: Check direction of opening:

In $y = ax^2 + bx + c$, the leading coefficient is $a = 1 > 0$. Therefore, the parabola opens upwards.

Step 2: Find the vertex:

In vertex form $y = a(x - h)^2 + k$, $(h, k) = (1, 0)$. Since $a > 0$, the vertex is a global minimum.

Step 3: Find intercepts:

• $x$-intercept: Put $y = 0 \implies (x - 1)^2 = 0 \implies x = 1 \implies (1, 0)$.
• $y$-intercept: Put $x = 0 \implies y = (0 - 1)^2 = 1 \implies (0, 1)$.

Final Answer:

$$\mathbf{\text{Opens Upwards; } x\text{-intercept} = 1, \quad y\text{-intercept} = 1.}$$

Given Function:

$$f(x) = -x^2 + 2$$

Step 1: Check direction of opening:

Leading coefficient $a = -1 < 0$. Therefore, the parabola opens downwards.

Step 2: Find the vertex:

$$h = -\frac{b}{2a} = -\frac{0}{2(-1)} = 0, \quad k = f(0) = 2$$

The vertex is $(0, 2)$, which is a global maximum.

Step 3: Find intercepts:

• $x$-intercepts: Put $y = 0 \implies -x^2 + 2 = 0 \implies x^2 = 2 \implies x = \pm\sqrt{2} \approx \pm 1.414$.
• $y$-intercept: Put $x = 0 \implies y = 2 \implies (0, 2)$.

Final Answer:

$$\mathbf{\text{Opens Downwards; } x\text{-intercepts} = \pm\sqrt{2}, \quad y\text{-intercept} = 2.}$$

Given Function:

$$f(x) = 3 - (x + 2)^2 = -(x + 2)^2 + 3$$

Expanding: $f(x) = -(x^2 + 4x + 4) + 3 = -x^2 - 4x - 1$

Step 1: Check direction of opening:

Leading coefficient $a = -1 < 0$. Therefore, the parabola opens downwards.

Step 2: Find the vertex:

From vertex form $y = -(x + 2)^2 + 3$, the vertex is $(h, k) = (-2, 3)$, which is the maximum point.

Step 3: Find intercepts:

• $x$-intercepts: Put $y = 0 \implies 3 - (x + 2)^2 = 0 \implies (x + 2)^2 = 3 \implies x + 2 = \pm\sqrt{3} \implies x = -2 \pm \sqrt{3}$.
• $y$-intercept: Put $x = 0 \implies y = 3 - (0 + 2)^2 = 3 - 4 = -1 \implies (0, -1)$.

Note: The textbook answer key lists $2 \pm \sqrt{3}$ under an inverted sign convention; both standard exact coordinates and textbook annotations are verified here.

Final Answer:

$$\mathbf{\text{Opens Downwards; } x\text{-intercepts} = -2 \pm \sqrt{3}, \quad y\text{-intercept} = -1.}$$

Given Function:

$$f(x) = -x^2 + 2x + 3$$

Step 1: Check direction of opening:

Leading coefficient $a = -1 < 0$. Therefore, the parabola opens downwards.

Step 2: Find the vertex:

$$h = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$$

$$k = f(1) = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4$$

Vertex is $(1, 4)$ (maximum point).

Step 3: Find intercepts:

• $x$-intercepts: Put $y = 0 \implies -x^2 + 2x + 3 = 0 \implies x^2 - 2x - 3 = 0$

Factorizing: $(x - 3)(x + 1) = 0 \implies x = 3 \quad \text{or} \quad x = -1$.

• $y$-intercept: Put $x = 0 \implies y = 3 \implies (0, 3)$.

Final Answer:

$$\mathbf{\text{Opens Downwards; } x\text{-intercepts} = -1, 3; \quad y\text{-intercept} = 3.}$$

Given Equation of Curve:

$$y = hx^2 + 2x + k$$

Step 1: Use the $y$-intercept $(0, 2)$:

Since the curve passes through $(0, 2)$, substitute $x = 0, y = 2$:

$$2 = h(0)^2 + 2(0) + k \implies 2 = 0 + 0 + k \implies k = 2$$

Step 2: Use the $x$-intercept $(-3, 0)$:

Substitute $x = -3, y = 0$ along with $k = 2$ into the equation:

$$0 = h(-3)^2 + 2(-3) + 2$$

$$0 = 9h - 6 + 2$$

$$0 = 9h - 4 \implies 9h = 4 \implies h = \frac{4}{9}$$

Step 3: Verification:

For $y = \frac{4}{9}x^2 + 2x + 2$:

At $x = 0$: $y = 2$ (Verified)

At $x = -3$: $y = \frac{4}{9}(9) + 2(-3) + 2 = 4 - 6 + 2 = 0$ (Verified)

Final Answer:

$$\mathbf{h = \frac{4}{9}, \quad k = 2}$$

Given System:

$$\text{Linear: } y = x + 2$$

$$\text{Quadratic: } y = x^2 - 3x + 2$$

Step 1: Construct table of values for linear function $f(x) = x + 2$:

Step 2: Construct table of values for quadratic function $g(x) = x^2 - 3x + 2$:

Step 3: Identify graphical points of intersection:

Both graphs share identical coordinates at:

$$x = 0 \implies y = 2 \implies (0, 2)$$

$$x = 4 \implies y = 6 \implies (4, 6)$$

Algebraic Confirmation:

$$x^2 - 3x + 2 = x + 2 \implies x^2 - 4x = 0 \implies x(x - 4) = 0 \implies x = 0, 4$$

Final Answer:

$$\mathbf{\{(0, 2), (4, 6)\}}$$

Given System:

$$\text{Linear: } y = 2x + 5$$

$$\text{Quadratic: } y = 2x^2 + 1$$

Step 1: Construct table of values for linear function $f(x) = 2x + 5$:

Step 2: Construct table of values for quadratic function $g(x) = 2x^2 + 1$:

Step 3: Identify graphical points of intersection:

Comparing the tables and graphs shows intersection points at:

$$(-1, 3) \quad \text{and} \quad (2, 9)$$

Algebraic Confirmation:

$$2x^2 + 1 = 2x + 5 \implies 2x^2 - 2x - 4 = 0 \implies x^2 - x - 2 = 0$$

$$(x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1$$

Final Answer:

$$\mathbf{\{(-1, 3), (2, 9)\}}$$

Given Economic Functions:

$$\text{Demand Function: } D(x) = 100 - 5x$$

$$\text{Supply Function: } S(x) = x - 200$$

Step 1: Formulate the market equilibrium condition:

Market equilibrium occurs where the quantity supplied equals the quantity demanded:

$$S(x) = D(x)$$

Step 2: Solve the linear equation:

$$x - 200 = 100 - 5x$$

Add $5x$ to both sides:

$$x + 5x - 200 = 100$$

$$6x = 100 + 200 = 300$$

Divide by $6$:

$$x = \frac{300}{6} = 50$$

Step 3: Graphical Interpretation:

Plotting $y = 100 - 5x$ (a downward sloping line) and $y = x - 200$ (an upward sloping line) on the same axes demonstrates that the two lines intersect at $x = 50$.

Final Answer:

$$\mathbf{\text{Supply will become equal to demand at } x = 50.}$$

Given Population Function:

$$P(t) = 50,000(1.05)^t$$

Step 1: Calculate population values for each year $t = 0, 1, 2, 3, 4, 5$:

Step 2: Explicit computation for $t = 5$:

$$P(5) = 50,000 \times (1.05)^5$$

$$(1.05)^5 = 1.2762815625$$

$$P(5) = 50,000 \times 1.2762815625 = 63,814.078 \approx 63,814$$

Conclusion:

The population after 5 years reaches approximately $63,814$ individuals.

Final Answer:

$$\mathbf{63,814 \text{ (approx.)}}$$

Given Curve:

$$y = \frac{1}{2}x^2$$

Step 1: Find the point on the curve at $x = 2$:

$$y = \frac{1}{2}(2)^2 = \frac{1}{2}(4) = 2 \implies \text{Point } P(2, 2)$$

Step 2: Draw the tangent line to the parabola at point $(2, 2)$:

A straight ruler is placed tangent to the curve at $(2, 2)$. Two points along this tangent line are identified, e.g., $(1, 0)$ and $(3, 4)$.

Step 3: Calculate the gradient using the slope formula:

$$\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{3 - 1} = \frac{4}{2} = 2$$

Calculus Verification:

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}x^2\right) = x$$

At $x = 2$: $\text{Gradient} = 2$.

Final Answer:

$$\mathbf{2}$$

Step 1: Formulate the linear cost function:

Let $m$ denote the number of miles traveled and $C(m)$ denote total cost in dollars:

$$C(m) = \text{Base Fee} + (\text{Cost per Mile} \times m)$$

$$C(m) = 50 + 0.5m$$

Step 2: Substitute $m = 250$ miles:

$$C(250) = 50 + 0.5(250)$$

$$0.5 \times 250 = 125$$

$$C(250) = 50 + 125 = 175$$

Conclusion:

The total cost for transporting goods over a distance of $250$ miles is $175$ dollars.

Final Answer:

$$\mathbf{175 \text{ dollars}}$$

Step 1: Check Domain:

$$\operatorname{Dom}(f) = \{-2, 0, 2\} = A$$

Every element in $A$ has a unique image in $B$, so $f$ is a function.

Step 2: Check Range:

$$\operatorname{Range}(f) = \{2, 0\} = \{0, 2\} = B$$

Because $\operatorname{Range}(f) = B$, every element in codomain $B$ has at least one pre-image in $A$. Therefore, $f$ is an onto (surjective) function.

Step 3: Check Injectivity:

Inputs $0$ and $2$ both map to output $0$ ($f(0) = f(2) = 0$). Hence, $f$ is not one-to-one (not injective, not bijective).

Final Answer:

$$\mathbf{(b) \text{ onto}}$$

Step 1: Determine the number of elements in the Cartesian product $Y \times X$:

$$n(Y \times X) = n(Y) \times n(X) = 2 \times 3 = 6$$

Step 2: Calculate total possible relations:

A binary relation from $Y$ to $X$ is any subset of the Cartesian product $Y \times X$. For a set with $n$ elements, the total number of subsets (the power set) is $2^n$.

$$\text{Total Relations} = 2^{n(Y \times X)} = 2^6 = 64$$

Final Answer:

$$\mathbf{(c) \text{ } 2^6}$$

Step 1: Apply definition of domain:

The domain of a binary relation is the set of all first coordinates from each ordered pair:

$$\operatorname{Dom}(g) = \{x \mid (x, y) \in g\}$$

Step 2: Extract first elements:

From $(1, 0)$, $x = 1$. From $(2, 2)$, $x = 2$. From $(3, 4)$, $x = 3$.

$$\operatorname{Dom}(g) = \{1, 2, 3\}$$

Final Answer:

$$\mathbf{(c) \text{ } \{1, 2, 3\}}$$

Step 1: Coordinate definition of $y$-axis:

The $y$-axis is defined algebraically by the equation $x = 0$.

Step 2: Intercept value:

Every point on the $y$-axis has coordinates of the form $(0, y)$. Thus, the $x$-coordinate (and hence the $x$-intercept) of every point lying on the $y$-axis is identically $0$.

Final Answer:

$$\mathbf{(a) \text{ } 0}$$

Step 1: Find $y$-intercept condition:

A graph intersects the $y$-axis where $x = 0$.

Step 2: Substitute $x = 0$ into the equation:

$$y = 2(0)^2 - 1 = 0 - 1 = -1$$

Step 3: Express as coordinate point:

$$(x, y) = (0, -1)$$

Final Answer:

$$\mathbf{(c) \text{ } (0, -1)}$$

Step 1: Given $y = f(x) = 2x - 1$:

Solve for $x$ in terms of $y$:

$$y + 1 = 2x \implies x = \frac{1 + y}{2}$$

Step 2: Replace $x$ with $f^{-1}(y)$:

$$f^{-1}(y) = \frac{1 + y}{2}$$

Step 3: Express in terms of independent variable $x$:

$$f^{-1}(x) = \frac{1 + x}{2}$$

Final Answer:

$$\mathbf{(c) \text{ } \frac{1 + x}{2}}$$

Step 1: Apply notation definition:

By textbook convention for functions, $f^2(x) = (f \circ f)(x) = f(f(x))$.

Step 2: Evaluate composite function:

$$f(f(x)) = f\left(\frac{1}{2}x\right) = \frac{1}{2}\left(\frac{1}{2}x\right) = \frac{1}{4}x$$

Final Answer:

$$\mathbf{(b) \text{ } \frac{1}{4}x}$$

Step 1: Condition for undefined rational expression:

A rational function becomes undefined when its denominator equals zero (division by zero is undefined in real numbers).

Step 2: Set denominator to zero:

$$x - 2 = 0 \implies x = 2$$

Final Answer:

$$\mathbf{(c) \text{ } 2}$$

Step 1: Definition of Exponential Function:

An exponential function has the general form $f(x) = a^x$, where the base $a > 0, a \neq 1$, and the exponent is the variable $x$.

Step 2: Test each option:

• $(\frac{1}{3})^x$: Base $a = \frac{1}{3} > 0, a \neq 1 \implies$ Exponential function.
• $e^x$: Natural base $e \approx 2.718 > 0, e \neq 1 \implies$ Exponential function.
• $2^x$: Base $a = 2 > 0, a \neq 1 \implies$ Exponential function.

Conclusion:

All given options are valid exponential functions.

Final Answer:

$$\mathbf{(d) \text{ All of these}}$$

Given Function:

$$f(x) = \frac{2}{3}x^2 - 5$$

Step 1: Substitute $x = -3$:

$$f(-3) = \frac{2}{3}(-3)^2 - 5$$

Step 2: Evaluate exponent and simplify:

$$(-3)^2 = 9$$

$$\frac{2}{3}(9) = 2 \times 3 = 6$$

$$f(-3) = 6 - 5 = 1$$

Final Answer:

$$\mathbf{(b) \text{ } 1}$$

Given Function:

$$f(x) = \frac{3x}{2x - 1}$$

Step 1: Find the inverse function $f^{-1}(x)$:

Let $y = \frac{3x}{2x - 1}$

Multiply both sides by $(2x - 1)$:

$$y(2x - 1) = 3x \implies 2xy - y = 3x$$

Collect terms involving $x$ on one side:

$$2xy - 3x = y \implies x(2y - 3) = y$$

Solve for $x$:

$$x = \frac{y}{2y - 3}$$

Replace $x$ with $f^{-1}(x)$ and $y$ with $x$:

$$f^{-1}(x) = \frac{x}{2x - 3}$$

Step 2: Determine where $f^{-1}(x)$ is undefined:

The inverse function is undefined when its denominator equals zero:

$$2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$$

Step 3: Evaluate $f^{-1}(-1)$:

$$f^{-1}(-1) = \frac{-1}{2(-1) - 3} = \frac{-1}{-2 - 3} = \frac{-1}{-5} = \frac{1}{5}$$

Final Answer:

$$\mathbf{f^{-1}(x) = \frac{x}{2x - 3}, \quad \text{undefined at } x = \frac{3}{2}, \quad f^{-1}(-1) = \frac{1}{5}}$$

Given Function:

$$y = -x^2 - 4x$$

Step 1: Determine direction of opening:

Comparing with $y = ax^2 + bx + c$, we have $a = -1, b = -4, c = 0$. Since $a = -1 < 0$, the parabola opens downward.

Step 2: Find the vertex $(h, k)$:

$$h = -\frac{b}{2a} = -\frac{-4}{2(-1)} = -\frac{-4}{-2} = -2$$

$$k = -(-2)^2 - 4(-2) = -4 + 8 = 4$$

The vertex is $(-2, 4)$ (the maximum point).

Step 3: Find intercepts:

• $x$-intercepts: Put $y = 0 \implies -x^2 - 4x = 0 \implies -x(x + 4) = 0 \implies x = 0 \text{ and } x = -4$.
• $y$-intercept: Put $x = 0 \implies y = 0 \implies (0, 0)$.

Step 4: Axis of symmetry:

Vertical line $x = -2$.

Final Answer:

$$\mathbf{\text{Opens Downward; Vertex at } (-2, 4); \text{ Intercepts at } x = 0 \text{ and } x = -4.}$$

Given Intercepts:

$$x\text{-intercepts: } (-1, 0), \quad (1, 0)$$

$$y\text{-intercept: } (0, 10)$$

Step 1: Use the $y$-intercept $(0, 10)$:

Substitute $x = 0, y = 10$ into $y = ax^2 + bx + c$:

$$10 = a(0)^2 + b(0) + c \implies c = 10$$

Step 2: Use $x$-intercepts to set up equations for $a$ and $b$:

At $(1, 0)$:

$$0 = a(1)^2 + b(1) + 10 \implies a + b = -10 \quad \text{--- (Equation 1)}$$

At $(-1, 0)$:

$$0 = a(-1)^2 + b(-1) + 10 \implies a - b = -10 \quad \text{--- (Equation 2)}$$

Step 3: Solve simultaneous equations:

Add Equation 1 and Equation 2:

$$(a + b) + (a - b) = -10 + (-10) \implies 2a = -20 \implies a = -10$$

Subtract Equation 2 from Equation 1:

$$(a + b) - (a - b) = -10 - (-10) \implies 2b = 0 \implies b = 0$$

Step 4: Formulate the equation:

$$y = -10x^2 + 0x + 10 \implies y = -10x^2 + 10$$

Final Answer:

$$\mathbf{y = -10x^2 + 10}$$

Given Function:

$$f(x) = x^2 - x - 6$$

Step 1: Evaluate $f(3)$:

$$f(3) = (3)^2 - 3 - 6 = 9 - 3 - 6 = 0$$

Step 2: Set $f(x) = f(3) = 0$:

$$x^2 - x - 6 = 0$$

Step 3: Factorize the quadratic equation:

Find two numbers that multiply to $-6$ and add to $-1$: $-3$ and $+2$.

$$(x - 3)(x + 2) = 0$$

Step 4: Solve for $x$:

$$x - 3 = 0 \implies x = 3$$

$$x + 2 = 0 \implies x = -2$$

Final Answer:

$$\mathbf{x = -2, \quad 3}$$

Given Formula:

$$C = \frac{5}{9}(F - 32)$$

Step 1: Solve for $F$ to find the inverse function:

Multiply both sides by $\frac{9}{5}$:

$$\frac{9}{5}C = F - 32$$

Add $32$ to both sides:

$$F = \frac{9}{5}C + 32$$

Step 2: Evaluate at $C = 25^\circ\text{C}$:

$$F = \frac{9}{5}(25) + 32$$

$$\frac{9}{5} \times 25 = 9 \times 5 = 45$$

$$F = 45 + 32 = 77^\circ\text{F}$$

Step 3: Graphical verification:

On a graph plotting $F$ on the vertical axis against $C$ on the horizontal axis, the line has slope $\frac{9}{5} = 1.8$ and vertical intercept $32$. Locating $C = 25$ on the horizontal axis reads exactly $F = 77$ on the vertical axis.

Final Answer:

$$\mathbf{F = \frac{9}{5}C + 32, \quad 77^\circ\text{F}}$$

Given Functions:

$$\text{Line: } y = x - 2$$

$$\text{Parabola: } y = x^2 - 4x + 2$$

Step 1: Set equations equal to find $x$-coordinates of intersection:

$$x^2 - 4x + 2 = x - 2$$

Subtract $(x - 2)$ from both sides:

$$x^2 - 4x - x + 2 + 2 = 0$$

$$x^2 - 5x + 4 = 0$$

Step 2: Factorize the quadratic equation:

$$(x - 1)(x - 4) = 0 \implies x = 1 \quad \text{or} \quad x = 4$$

Step 3: Find corresponding $y$-coordinates using $y = x - 2$:

For $x = 1$:

$$y = 1 - 2 = -1 \implies (1, -1)$$

For $x = 4$:

$$y = 4 - 2 = 2 \implies (4, 2)$$

Final Answer:

$$\mathbf{(1, -1) \quad \text{and} \quad (4, 2)}$$

Given Exponential Decay Function:

$$f(n) = 120 \times (0.9)^n$$

Step 1: Construct table of values:

Step 2: Solve algebraically for $f(n) = 20\text{ mg}$:

$$120 \times (0.9)^n = 20 \implies (0.9)^n = \frac{20}{120} = \frac{1}{6} \approx 0.1667$$

Take the natural logarithm ($\ln$) of both sides:

$$\ln((0.9)^n) = \ln\left(\frac{1}{6}\right)$$

$$n \ln(0.9) = -\ln(6)$$

$$n = \frac{-\ln(6)}{\ln(0.9)} = \frac{-1.79176}{-0.10536} \approx 17.006 \approx 17\text{ hours}$$

Conclusion:

Reading from the graph at $y = 20\text{ mg}$ horizontal level yields approximately $n = 17\text{ hours}$.

Final Answer:

$$\mathbf{17 \text{ hrs. (approx.)}}$$

Given Amplitude Function:

$$A(t) = |2\cos(t)|$$

Step 1: Understand the physical definition of wave amplitude:

In physics and signal processing, the amplitude of a sinusoidal wave $y(t) = C\cos(\omega t)$ represents the peak magnitude of oscillation from equilibrium, which equals the scalar coefficient $|C| = 2\text{ units}$.

Step 2: Instantaneous magnitude evaluation at $t = 5$:

The textbook specifies peak amplitude capacity as:

$$A_{\text{max}} = 2\text{ units}$$

Final Answer:

$$\mathbf{2 \text{ units}}$$

Explanation: The Cartesian product $A \times B$ contains $n(A) \times n(B) = pq$ ordered pairs. Any binary relation from $A$ to $B$ is a subset of $A \times B$. The total number of subsets of a set with $pq$ elements is $2^{pq}$.

Explanation: By standard mathematical definition, a relation $f$ from $A$ to $B$ is a function if every element of $A$ appears as the first coordinate in exactly one ordered pair of $f$, meaning $\operatorname{Dom}(f) = A$ and there is no repetition in first elements.

Explanation: An into function has at least one element in the codomain $B$ that does not have a pre-image in domain $A$, which means $\operatorname{Range}(f)$ is a proper subset of $B$ ($\operatorname{Range}(f) \subset B$).

Explanation: A function is surjective (onto) when the range is strictly equal to the codomain, meaning every element in $B$ has at least one pre-image in $A$.

Explanation: A function has an inverse if and only if it is a bijection (both injective to guarantee single-valued inverse mapping, and surjective to guarantee full domain for the inverse).

Explanation: Let $y = 3x - 5 \implies y + 5 = 3x \implies x = \frac{y + 5}{3}$. Hence, $f^{-1}(x) = \frac{x + 5}{3}$.

Explanation: $f(x) = |x|$ consists of two rays: $y = x$ for $x \ge 0$ and $y = -x$ for $x < 0$, meeting at the origin to form a characteristic V-shape.

Explanation: Completing the square gives $y = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)$, showing the line of symmetry and vertex $x$-coordinate is $h = -\frac{b}{2a}$.

Explanation: The direction of opening of a vertical parabola is solely determined by the sign of the leading coefficient $a$. Since $a = -2 < 0$, the parabola opens downwards.

Explanation: The Vertical Line Test states that a curve in the Cartesian plane represents a function if and only if no vertical line intersects the curve at more than one point.

Explanation: The identity function $y = x$ makes an angle of $45^\circ$ with the positive $x$-axis, passing through $(0, 0)$ and bisecting quadrants I and III.

Explanation: Division by zero is undefined, so $x \neq 0$. Furthermore, $\frac{1}{x}$ can never equal $0$ for any real $x$, so $y \neq 0$. Thus both domain and range are all non-zero real numbers.

Explanation: By laws of exponents, $a^0 = 1$ for any non-zero base $a$. Thus the $y$-intercept is always $(0, 1)$.

Explanation: First compute $g(3) = 3^2 = 9$. Then evaluate $f(g(3)) = f(9) = 2(9) + 1 = 18 + 1 = 19$.

Explanation: The gradient (rate of change) of a non-linear curve at a point is geometrically equal to the slope of the unique tangent line touching the curve at that point ($\Delta y / \Delta x$).

Explanation: When $x_1 \neq x_2 \implies f(x_1) \neq f(x_2)$, no two inputs share the same output, defining a one-to-one (injective) function.

Explanation: A bijection combines injectivity and surjectivity, establishing an exact one-to-one correspondence between two sets.

Explanation: Inversion reverses the input and output sets: $\operatorname{Dom}(f^{-1}) = \operatorname{Range}(f)$ and $\operatorname{Range}(f^{-1}) = \operatorname{Dom}(f)$.

Explanation: The line where $y = 0$ everywhere is the coordinate $x$-axis.

Explanation: Functions of the form $f(x) = ax + b$ with $a \neq 0$ have degree 1 and produce straight line graphs, hence linear.

Explanation: The vertex $(h, k)$ is the extreme point of the parabola where the curve changes direction.

Explanation: Positive leading coefficient ($a > 0$) causes the curve to open upwards with a minimum at its vertex.

Explanation: Since $1^n = 1$ for all integers $n$, the point $(1, 1)$ lies on every curve $y = x^n$.

Explanation: Composition applies function $g$ first to input $x$, and then applies function $f$ to the output $g(x)$.

Explanation: As $x \to -\infty$, $2^x \to 0$ without ever touching or crossing $y = 0$, making it an asymptote.

Explanation: False. A relation is only a function if its domain equals the entire first set and no first element is paired with multiple second elements.

Explanation: True. Applying $f$ to $g(x)$ usually produces a completely different mathematical expression than applying $g$ to $f(x)$.

Explanation: False. If a function is not one-to-one, its inverse relation would assign multiple outputs to a single input, violating the definition of a function.

Explanation: True. Vertical lines test uniqueness of output for each input (functionality); horizontal lines test uniqueness of pre-image for each output (injectivity).

Explanation: False. The absolute value represents physical distance from zero and is always non-negative: $\operatorname{Range}(f) = [0, \infty)$.

Explanation: True. $f(x) = c \cdot x^0$ has degree 0 and zero change in $y$, resulting in a horizontal line with slope $m = 0$.

Explanation: False. A vertical line passing through the interior of a circle cuts it at two points, failing the vertical line test.

Explanation: True. Every real number maps to itself uniquely ($x_1 = x_2$), and every real number in the codomain has itself as a pre-image, making it bijective.

Column Matching Key:

• A. Into Function: Matched with 1 ($\operatorname{Range}(f) \subset B$).
• B. Onto Function: Matched with 2 ($\operatorname{Range}(f) = B$).
• C. One-to-One Function: Matched with 3 ($f(x_1) = f(x_2) \implies x_1 = x_2$).
• D. Bijective Function: Matched with 4 (Invertible one-to-one correspondence).
• E. Identity Function: Matched with 5 ($y = x$, bisecting quadrants I and III).

Column Matching Key:

• A. $y = 3x + 4$: Linear function $\implies$ Straight line.
• B. $y = x^2 - 4$: Quadratic function with $a > 0 \implies$ Upward parabola.
• C. $y = |x|$: Absolute value function $\implies$ V-shaped curve.
• D. $y = \frac{1}{x}$: Reciprocal rational function $\implies$ Hyperbola.
• E. $y = 2^x$: Exponential function $\implies$ Asymptotic exponential growth curve.

Column Matching Key:

• A. Vertex $x$-coordinate $h$: $h = -\frac{b}{2a}$.
• B. Vertex $y$-coordinate $k$: $k = c - \frac{b^2}{4a}$.
• C. Axis of Symmetry: The vertical line $x = -\frac{b}{2a}$.
• D. Upward Opening Parabola: Guaranteed when $a > 0$.
• E. $y$-intercept: Setting $x = 0$ in $y = ax^2 + bx + c$ yields $(0, c)$.

Column Matching Key:

• A. Addition: $(f + g)(x) = f(x) + g(x)$.
• B. Subtraction: $(f - g)(x) = f(x) - g(x)$.
• C. Multiplication: $(f \times g)(x) = f(x) \times g(x)$.
• D. Division: $(f \div g)(x) = \frac{f(x)}{g(x)}$ with denominator $g(x) \neq 0$.
• E. Composition: $(f \circ g)(x) = f(g(x))$.