Problem 1: Refer to $g(x) = 7x + 2$

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(a) Domain and Range of $g(x)$:

• Domain: The domain of $g(x)$ is all real numbers ($\mathbb{R}$), since there are no restrictions on $x$.
• Range: The function $g(x)$ is a linear function with a non-zero slope ($7$), so its range is also all real numbers ($\mathbb{R}$).

(b) Algebraic Expression for $g(h(x))$:

Let $h(x) = x^2 + x + 1$. Substituting $h(x)$ into $g(x)$: $g(h(x)) = 7h(x) + 2 = 7(x^2 + x + 1) + 2 = 7x^2 + 7x + 7 + 2 = 7x^2 + 7x + 9$

(c) Algebraic Expression for $3g(x) \circ h(x)$:

We need $3g(x) \circ h(x)$, which implies: $3g(x) \circ h(x) = 3g(h(x)) = 3(7x^2 + 7x + 9) = 21x^2 + 21x + 27$

(d) Function Equation for the Inverse Function $g^{-1}(x)$:

To find $g^{-1}(x)$, solve for $x$ in $y = 7x + 2$: $y = 7x + 2 \implies y - 2 = 7x \implies x = \frac{y - 2}{7}$ Thus, the inverse function is: $g^{-1}(x) = \frac{x - 2}{7}$

(e) Domain and Range of $g^{-1}(x)$:

• Domain: Since the range of $g(x)$ is all real numbers, the domain of $g^{-1}(x)$ is also $\mathbb{R}$.
• Range: The range of $g^{-1}(x)$ is the domain of $g(x)$, which is $\mathbb{R}$.

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Problem 2: Refer to the graph of $R(x)$

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(a) Domain and Range of $R(x)$:

From the graph, the $x$-values span from $-7.5$ to $6$, and the $y$-values span from $0$ to $6$.

• Domain: $[-7.5, 6]$
• Range: $[0, 6]$

(b) Graph of the Inverse Function $R^{-1}(x)$:

The inverse function graph $R^{-1}(x)$ is obtained by reflecting the points of $R(x)$ across the line $y = x$. The points for $R(x)$ become: $(-7.5, 0) \rightarrow (0, -7.5), \, (-1, 2) \rightarrow (2, -1), \, (0, 4.5) \rightarrow (4.5, 0), \, (6, 6) \rightarrow (6, 6)$ Plot these points to obtain $R^{-1}(x)$.

(c) Domain and Range of $R^{-1}(x)$:

• Domain: Since the range of $R(x)$ is $[0, 6]$, the domain of $R^{-1}(x)$ is $[0, 6]$.
• Range: Since the domain of $R(x)$ is $[-7.5, 6]$, the range of $R^{-1}(x)$ is $[-7.5, 6]$.

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Problem 3: Refer to $f(x)$ in Table 3.14

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(a) Domain and Range of $f(x)$:

From the table:

• Domain: The $x$-values are $\{0, 1, 2, 3\}$.
• Range: The $f(x)$-values are $\{3, 0, 2, -4\}$.

(b) Function Table for the Inverse Function $f^{-1}(x)$:

To find $f^{-1}(x)$, swap the domain and range of $f(x)$: $f(x) = \{ (0, 3), (1, 0), (2, 2), (3, -4) \} \implies f^{-1}(x) = \{ (3, 0), (0, 1), (2, 2), (-4, 3) \}$

(c) Domain and Range of $f^{-1}(x)$:

• Domain: The range of $f(x)$, which is $\{3, 0, 2, -4\}$.
• Range: The domain of $f(x)$, which is $\{0, 1, 2, 3\}$.

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Follow-Up Questions:

1. How can we verify the correctness of the inverse functions derived in each case?
2. What is the relationship between the slopes of a linear function and its inverse?
3. How does the reflection property $y = x$ affect the graphical representation of an inverse function?
4. Can the composition of $g$ and $g^{-1}$ demonstrate their inverse relationship? Show examples.
5. What happens to the domain and range if $g(x) = 7x + 2$ were modified to a quadratic function?

Tip:

To check if two functions are inverses, calculate their compositions ($f(g(x))$ and $g(f(x))$); if both equal $x$, they are inverses.