We are given a function $f: D \to C$, defined as $f(x) = |2x - 3|$, with $D, C \subseteq \mathbb{R}$. We need to provide examples for $D$ and $C$ where $f$ is injective, not injective, surjective, and not surjective.

(a) Injectivity

A function is injective if different inputs lead to different outputs, i.e., if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Case 1: $f$ is injective

To ensure injectivity, we must choose a domain $D$ where the function $f(x) = |2x - 3|$ is either strictly increasing or strictly decreasing. This happens when we restrict the domain to one side of the vertex $x = \frac{3}{2}$ because the absolute value function will behave monotonically.

Example: Let $D = [\frac{3}{2}, \infty)$ and $C = [0, \infty)$. For $x \in [\frac{3}{2}, \infty)$, the function $f(x) = 2x - 3$ (no absolute value, since $2x - 3 \geq 0$). This is a linear and strictly increasing function, so $f$ is injective on this domain.

Case 2: $f$ is not injective

If the domain includes values on both sides of $x = \frac{3}{2}$, the function will no longer be injective because the absolute value will lead to two different inputs producing the same output.

Example: Let $D = \mathbb{R}$ and $C = [0, \infty)$. For instance, $f(0) = |2(0) - 3| = 3$ and $f(3) = |2(3) - 3| = 3$, so $f(0) = f(3)$, which violates injectivity.

(b) Surjectivity

A function is surjective if every element in the codomain $C$ has a preimage in the domain $D$, i.e., for every $y \in C$, there exists an $x \in D$ such that $f(x) = y$.

Case 1: $f$ is surjective

To ensure surjectivity, we need to choose the codomain $C$ such that every value $f(x) = |2x - 3|$ can be achieved by some $x \in D$. The minimum value of $f(x)$ is 0, which occurs at $x = \frac{3}{2}$. Therefore, for surjectivity, we can choose $C = [0, \infty)$.

Example: Let $D = \mathbb{R}$ and $C = [0, \infty)$. For every $y \geq 0$, there exists some $x \in \mathbb{R}$ such that $f(x) = y$. For instance, solving $|2x - 3| = y$ gives two possible solutions $x = \frac{y + 3}{2}$ or $x = \frac{-y + 3}{2}$, ensuring that all $y \geq 0$ are covered.

Case 2: $f$ is not surjective

If the codomain $C$ includes values that are negative, $f$ will not be surjective, as the absolute value function only takes non-negative values.

Example: Let $D = \mathbb{R}$ and $C = \mathbb{R}$. The function $f(x) = |2x - 3|$ cannot produce any negative outputs, so $f$ is not surjective onto $\mathbb{R}$.

Would you like more detailed steps or clarifications? Here are 5 related questions to explore further:

1. How would the injectivity of the function change if the definition of $f$ used $2x + 3$ instead of $2x - 3$?
2. What happens to surjectivity if we change the domain $D$ to a subset of $\mathbb{R}$?
3. Can you generalize conditions for injectivity and surjectivity for any absolute value function of the form $|ax - b|$?
4. How does the function's behavior change when we choose $D = [0, 2]$?
5. What are some graphical methods to visualize injectivity and surjectivity?

Tip: When working with absolute value functions, it's helpful to break the problem into cases: one where the expression inside the absolute value is positive and another where it's negative.