Step 1: Establishing the Governing Equation

The relationship between temperature in degrees Celsius ($C$) and degrees Fahrenheit ($F$) is provided by the linear function:

$t(C) = \frac{9}{5}C + 32$

where $t(C)$ represents the temperature in degrees Fahrenheit. [Per the definition of the linear temperature conversion function].

Step 2: Substituting the Known Variable

We are given that the temperature in Fahrenheit is $t(C) = 212$. We substitute this value into the governing equation to solve for the unknown variable $C$:

$212 = \frac{9}{5}C + 32$

Step 3: Algebraic Isolation of the Variable $C$

To isolate the term containing $C$, we apply the subtraction property of equality by subtracting 32 from both sides of the equation:

$212 - 32 = \frac{9}{5}C$

$180 = \frac{9}{5}C$

Step 4: Solving for $C$

To isolate $C$, multiply both sides of the equation by the reciprocal of the coefficient $\frac{9}{5}$, which is $\frac{5}{9}$:

$C = 180 \times \frac{5}{9}$

$C = \left(\frac{180}{9}\right) \times 5$ [By the Associative Property of Multiplication]

$C = 20 \times 5$

$C = 100$

Visual Representation of the Linear Relationship

Final Solution: The value of $C$ when $t(C) = 212$ is $100^{\circ}\text{C}$.

Step 1: Theoretical Definition of a Function

A relation $f$ from a set $A$ to a set $B$ is defined as a function if and only if every element in the domain $A$ has exactly one unique image in the codomain $B$. [Per the definition of a mapping in Set Theory].

Step 2: Analysis of the Given Relation

The given relation $R$ is defined as the set of ordered pairs: $R = \{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)\}$.

To determine if $R$ is a function, we map each element of the domain ($x$) to its corresponding image ($y$):

• $x = 2 \implies f(2) = 1$
• $x = 4 \implies f(4) = 2$
• $x = 6 \implies f(6) = 3$
• $x = 8 \implies f(8) = 4$
• $x = 10 \implies f(10) = 5$
• $x = 12 \implies f(12) = 6$
• $x = 14 \implies f(14) = 7$

Observation: Each distinct input $x$ in the domain set $\{2, 4, 6, 8, 10, 12, 14\}$ corresponds to exactly one unique output $y$ in the range set $\{1, 2, 3, 4, 5, 6, 7\}$. [By the Vertical Line Test and Function Mapping Principle]. Therefore, $R$ is a function.

Step 3: Visualization of the Mapping

Step 4: Determination of Domain and Range

Final Solution: The relation is a function because every element of the domain has a unique image in the range. The domain is $\{2, 4, 6, 8, 10, 12, 14\}$ and the range is $\{1, 2, 3, 4, 5, 6, 7\}$.

Step 1: Theoretical Definition of a Function

A relation $f$ from a set $A$ to a set $B$ is defined as a function if every element $x \in A$ has one and only one image $y \in B$. [Per the Definition of a Mapping/Function]. Specifically, for a set of ordered pairs, a relation is a function if and only if no two distinct ordered pairs share the same first element (domain value).

Step 2: Analysis of the Given Relation

Let the given relation be $R = \{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)\}$.

Observation: Every input value ($2, 5, 8, 11, 14, 17$) is mapped to exactly one output value ($1$). There are no instances where the same input yields different outputs. [By the Uniqueness Condition of Functions].

Step 3: Determination of Domain and Range

• Domain: The set of all first coordinates in the ordered pairs.
  $\text{Domain} = \{2, 5, 8, 11, 14, 17\}$
• Range: The set of all second coordinates in the ordered pairs.
  $\text{Range} = \{1\}$

Step 4: Visual Representation

Final Solution: The relation is a function because every element in the domain has a unique image in the range. The domain is $\{2, 5, 8, 11, 14, 17\}$ and the range is $\{1\}$.

Step 1: Formal Definition of the Function and Domain

The given function is defined as $f(x) = 2 - 3x$. The domain of the function is restricted by the condition $x \in \mathbb{R}$ and $x > 0$. In interval notation, the domain is $(0, \infty)$.

Step 2: Analytical Transformation of the Inequality

To determine the range, we must find the set of all possible values for $f(x)$ as $x$ varies across its domain $(0, \infty)$. We begin with the given inequality [By Definition of Domain]:

$x > 0$

Multiply both sides by $-3$. Per the rules of inequality, multiplying by a negative number reverses the inequality sign [Property of Inequalities]:

$-3x < -3(0)$

$-3x < 0$

Step 3: Completing the Expression

To reach the functional form $f(x) = 2 - 3x$, add $2$ to both sides of the inequality [Property of Addition in Inequalities]:

$2 - 3x < 2 + 0$

$2 - 3x < 2$

Since $f(x) = 2 - 3x$, we substitute the function notation:

$f(x) < 2$

Step 4: Mapping the Range

As $x$ approaches $0$ from the positive side, $f(x)$ approaches $2 - 3(0) = 2$. As $x \to \infty$, $f(x)$ decreases infinitely toward $-\infty$. Therefore, the range is the set of all values strictly less than $2$.

Step 5: Final Set Representation

The range represents all real numbers $y$ such that $y < 2$. In set-builder notation and interval notation:

• Set-builder notation: $\{y \in \mathbb{R} : y < 2\}$
• Interval notation: $(-\infty, 2)$

Final Solution: The range of the function $f(x) = 2 - 3x$ for $x > 0$ is $(-\infty, 2)$.

Step 1: Theoretical Foundation of Function Evaluation

A function $f: A \rightarrow B$ assigns to each element $x \in A$ exactly one element $f(x) \in B$. To evaluate a function $f(x)$ for a specific value $x = a$, we perform a direct substitution of the constant $a$ into the variable $x$ within the defined algebraic rule [Per the definition of functional mapping].

Step 2: Substitution of Given Variable

The given function is defined as:

$f(x) = 2x - 5$

We are required to find the value of $f(0)$. Here, the independent variable $x$ is assigned the value of $0$.

$f(0) = 2(0) - 5$ [By substitution property of equality]

Step 3: Algebraic Simplification

Applying the arithmetic operations as per the order of operations (BODMAS/PEMDAS):

• Multiply the coefficient $2$ by the input value $0$: $2 \times 0 = 0$ [By the Multiplicative Property of Zero].
• Subtract the constant $5$ from the product: $0 - 5 = -5$ [By the Identity Property of Subtraction].

Visual Representation of the Linear Function

The graph represents the line $y = 2x - 5$. At $x = 0$, the line intersects the y-axis at the point $(0, -5)$, which confirms our analytical calculation.

Final Solution: $f(0) = -5$

Step 1: Establishing the Functional Mapping

The function $t: \mathbb{R} \to \mathbb{R}$ is defined by the linear transformation equation representing the conversion from the Celsius scale to the Fahrenheit scale:

$t(C) = \frac{9C}{5} + 32$

Where $C$ represents the input variable (temperature in degrees Celsius) and $t(C)$ represents the output (temperature in degrees Fahrenheit).

Step 2: Substitution of the Independent Variable

To find the value of $t(-10)$, we substitute $C = -10$ into the defined function. [By the Principle of Function Evaluation]:

$t(-10) = \frac{9(-10)}{5} + 32$

Step 3: Algebraic Simplification

Following the Order of Operations (PEMDAS/BODMAS):

• Perform the multiplication in the numerator: $9 \times (-10) = -90$
• Divide the result by the denominator: $\frac{-90}{5} = -18$
• Add the constant term: $-18 + 32$

Step 4: Final Calculation

Applying the arithmetic sum:

$t(-10) = -18 + 32 = 14$

Visual Representation of Linear Mapping

Final Solution: $t(-10) = 14$ degrees Fahrenheit.

Step 1: Analyzing the Function Definition

The function is defined as $f(x) = -|x|$, where $x \in \mathbb{R}$. The absolute value function $|x|$ is defined as:

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

By applying the negative sign, the function becomes:

$f(x) = \begin{cases} -x, & \text{if } x \ge 0 \\ x, & \text{if } x < 0 \end{cases}$

Step 2: Determining the Domain

The domain of a function $f(x)$ is the set of all possible input values ($x$) for which the function is defined. [By Definition of Real Function]. Since $|x|$ is defined for all real numbers $x \in \mathbb{R}$, and multiplying by $-1$ does not restrict the input set, the function $f(x) = -|x|$ is well-defined for all real numbers.

Domain of $f$ = $(-\infty, \infty)$ or $\mathbb{R}$.

Step 3: Determining the Range

The range is the set of all output values $y = f(x)$.

• Since the absolute value $|x| \ge 0$ for all $x \in \mathbb{R}$ [Property of Absolute Value], it follows that $-|x| \le 0$ for all $x \in \mathbb{R}$.
• As $x \to \infty$ or $x \to -\infty$, $f(x) \to -\infty$.
• When $x = 0$, $f(0) = -|0| = 0$.

Range of $f$ = $(-\infty, 0]$.

Step 4: Visual Representation

Step 5: Tabular Summary

Final Solution: The domain of $f(x) = -|x|$ is $\mathbb{R}$ and the range is $(-\infty, 0]$.

Step 1: Identifying the Domain Condition

For the function $f(x) = \sqrt{9-x^2}$ to be defined as a real-valued function, the expression inside the square root (the radicand) must be non-negative [Per the definition of square root functions in the set of real numbers $\mathbb{R}$].

Thus, we set the following inequality:

$9 - x^2 \geq 0$

Step 2: Solving for x

Rearranging the inequality:

$x^2 \leq 9$

By taking the square root on both sides and considering the absolute value property $|x| \leq a \iff -a \leq x \leq a$:

$\sqrt{x^2} \leq \sqrt{9}$

$|x| \leq 3$

$-3 \leq x \leq 3$

Therefore, the domain of the function is the interval $[-3, 3]$.

Step 3: Determining the Range

Since $f(x) = \sqrt{9-x^2}$, we analyze the bounds of the function based on the domain determined in Step 2.

• Minimum value: When $|x| = 3$, $x^2 = 9$. Then, $f(3) = \sqrt{9-9} = 0$.
• Maximum value: When $x = 0$, $f(0) = \sqrt{9-0} = \sqrt{9} = 3$.

Since the square root function $\sqrt{u}$ is monotonically increasing for $u \geq 0$, the range of $f(x)$ spans from the absolute minimum to the absolute maximum.

Thus, the range is $[0, 3]$.

Visual Representation of the Function

Analytical Summary

Final Solution: The domain of $f(x) = \sqrt{9-x^2}$ is $[-3, 3]$ and the range is $[0, 3]$.

1. Definitions and Theoretical Foundation

The function is defined as $f(x) = x$, where $x \in \mathbb{R}$. This is classified as the Identity Function. By definition, the identity function $f: \mathbb{R} \to \mathbb{R}$ maps every real number to itself. [Per the definition of the Identity Function, $f(x) = x$ for all $x$ in the domain].

2. Domain Analysis

The domain of the function $f$ is explicitly stated as the set of all real numbers.
Domain of $f = \{x : x \in \mathbb{R}\}$ or $(-\infty, \infty)$.

3. Range Determination

The range is defined as the set of all output values $f(x)$ for every $x$ in the domain.
Since $f(x) = x$:

• If $x = 1$, $f(1) = 1$
• If $x = -5.5$, $f(-5.5) = -5.5$
• If $x = 0$, $f(0) = 0$

4. Visual Representation

5. Tabular Summary of Mapping

Final Solution: The range of the function $f(x) = x$ is the set of all real numbers, denoted as $\mathbb{R}$ or $(-\infty, \infty)$.

Step 1: Formal Definition of a Function

A relation $f: A \rightarrow B$ is defined as a function if and only if for every element $x$ in the domain $A$, there exists one and only one image $y$ in the codomain $B$ [Per the definition of a mapping in set theory]. If any single input $x$ is associated with more than one output $y$, the relation fails the vertical line test and is not a function.

Step 2: Analysis of the Given Relation

The given relation is $R = \{(1, 3), (1, 5), (2, 5)\}$.

To determine if $R$ is a function, we must examine the mapping of the elements in the domain:

• For input $x = 1$, the relation maps to two distinct outputs: $y = 3$ and $y = 5$.
• For input $x = 2$, the relation maps to the output: $y = 5$.

Since the input $1$ is associated with two different outputs ($3 \neq 5$), the uniqueness criterion of a function is violated [By the definition of a well-defined function].

Step 3: Visual Representation

Step 4: Conclusion

Because the input $1$ produces two different outputs ($3$ and $5$), the relation $R$ is not a function.

Final Solution: The relation $\{(1,3), (1,5), (2,5)\}$ is not a function because the element $1$ in the domain has two distinct images ($3$ and $5$).

Step 1: Establishing the Functional Definition

The function $t(C)$ represents the mapping from Celsius ($C$) to Fahrenheit ($F$). The given mathematical relationship is:

$t(C) = \frac{9}{5}C + 32$

[By the definition of a linear function, where the slope $m = 1.8$ and the y-intercept $b = 32$].

Step 2: Substitution of the Independent Variable

To evaluate the function at $C = 0$, we substitute the value $0$ for the variable $C$ in the provided expression:

$t(0) = \frac{9(0)}{5} + 32$

Step 3: Algebraic Computation

Applying the properties of multiplication by zero:

• $\frac{9 \times 0}{5} = \frac{0}{5} = 0$ [By the Zero Property of Multiplication]
• $t(0) = 0 + 32$ [By the Additive Identity Property]
• $t(0) = 32$

Visual Representation of the Linear Function

Final Solution: $t(0) = 32$

Step 1: Establishing the Functional Mapping

The function $t(C)$ represents a linear transformation from the Celsius scale ($C$) to the Fahrenheit scale ($F$). The given definition is:

$t(C) = \frac{9C}{5} + 32$

[Per the definition of a linear function in temperature conversion protocols]

Step 2: Substitution of the Independent Variable

To determine $t(28)$, we substitute the value of the independent variable $C = 28$ into the function:

$t(28) = \frac{9(28)}{5} + 32$

Step 3: Algebraic Computation

Performing the multiplication in the numerator:

$9 \times 28 = 252$

$t(28) = \frac{252}{5} + 32$

[By the properties of arithmetic operations on real numbers]

Step 4: Division and Final Summation

Calculate the quotient of $252$ divided by $5$:

$\frac{252}{5} = 50.4$

Now, add the constant term of the function:

$t(28) = 50.4 + 32 = 82.4$

Visual Representation of Linear Mapping

Final Solution: $t(28) = 82.4$

Step 1: Identifying the Functional Definition

The given function is defined as a mapping $f: \mathbb{R} \to \mathbb{R}$ such that the rule for the transformation of any element $x$ in the domain is:

$f(x) = 2x - 5$

[Definition of a linear function in terms of its algebraic expression]

Step 2: Substitution of the Independent Variable

To determine the value of the function at $x = 7$, we substitute the value $7$ into the expression for $f(x)$.

$f(7) = 2(7) - 5$

[Substitution Property of Equality]

Step 3: Arithmetic Evaluation

Perform the operations following the Order of Operations (PEMDAS/BODMAS):

• Multiply the scalar coefficient by the input value: $2 \times 7 = 14$
• Subtract the constant term: $14 - 5 = 9$

Step 4: Final Verification

The calculation $f(7) = 2(7) - 5 = 14 - 5 = 9$ confirms that the image of the element $7$ under the function $f$ is $9$.

Final Solution: $f(7) = 9$

Step 1: Identifying the Function Definition

The function is explicitly defined as a linear mapping $f: \mathbb{R} \to \mathbb{R}$ given by the algebraic expression:

$f(x) = 2x - 5$ [Given functional rule]

Step 2: Substitution of the Independent Variable

To determine the value of the function at a specific point $x = c$, we substitute the value $c$ into the variable $x$ within the expression. For the requested value $f(-3)$, we set $x = -3$.

$f(-3) = 2(-3) - 5$ [By the principle of functional substitution]

Step 3: Algebraic Evaluation

Perform the multiplication followed by the subtraction as dictated by the order of operations (PEMDAS/BODMAS):

• First, calculate the product: $2 \times (-3) = -6$ [By the property of multiplication of integers]
• Next, evaluate the final expression: $-6 - 5 = -11$ [By the property of subtraction of signed numbers]

Visual Representation of the Linear Function

Summary of Calculation

Final Solution: $f(-3) = -11$

Step 1: Identifying the Domain and Function Definition

Given the function $f(x) = x^2 + 2$, where $x \in \mathbb{R}$ (the set of all real numbers). The domain is defined as the set of all possible input values for $x$, which is $(-\infty, \infty)$.

Step 2: Analyzing the Square Property of Real Numbers

By the definition of real numbers, the square of any real number is always non-negative [Axiom of non-negativity in ordered fields]:

For all $x \in \mathbb{R}$, $x^2 \geq 0$.

Step 3: Transforming the Inequality to Match the Function

To determine the range, we must transform $x^2$ into $x^2 + 2$ by applying the Addition Property of Inequalities [Property of Ordered Fields]:

Add $2$ to both sides of the inequality $x^2 \geq 0$:

$x^2 + 2 \geq 0 + 2$

$x^2 + 2 \geq 2$

Step 4: Establishing the Range

Since $f(x) = x^2 + 2$, we conclude that $f(x) \geq 2$. The range is the set of all output values $y$ such that $y = f(x)$. Therefore, the range of the function is the interval $[2, \infty)$.

Final Solution: The range of the function $f(x) = x^2 + 2$ is $[2, \infty)$.