The floor function $\lfloor 5 - 2x \rfloor$ gives the greatest integer less than or equal to $5 - 2x$. To express $f(x) = \lfloor 5 - 2x \rfloor$ as a piecewise function, we need to break it down into intervals where the value of $5 - 2x$ lies between successive integers, as the floor function will output a constant integer on each such interval.

We examine intervals where $5 - 2x$ takes integer values:

1. For $5 - 2x \in [n, n+1)$, where $n$ is an integer, $\lfloor 5 - 2x \rfloor = n$.

Let’s find the intervals by solving for $x$ in terms of $5 - 2x = n$.

1. When $5 - 2x \in [4, 5)$, then $\lfloor 5 - 2x \rfloor = 4$:

$4 \leq 5 - 2x < 5$ Solving for $x$: $4 \leq 5 - 2x \implies x \leq \frac{1}{2}$ $5 - 2x < 5 \implies x > 0$ Thus, for $0 < x \leq \frac{1}{2}$, $f(x) = 4$.

2. When $5 - 2x \in [3, 4)$, then $\lfloor 5 - 2x \rfloor = 3$:

$3 \leq 5 - 2x < 4$ Solving for $x$: $3 \leq 5 - 2x \implies x \geq \frac{1}{2}$ $5 - 2x < 4 \implies x < 1$ Thus, for $\frac{1}{2} \leq x < 1$, $f(x) = 3$.

3. When $5 - 2x \in [2, 3)$, then $\lfloor 5 - 2x \rfloor = 2$:

$2 \leq 5 - 2x < 3$ Solving for $x$: $2 \leq 5 - 2x \implies x \geq 1$ $5 - 2x < 3 \implies x < \frac{3}{2}$ Thus, for $1 \leq x < \frac{3}{2}$, $f(x) = 2$.

4. When $5 - 2x \in [1, 2)$, then $\lfloor 5 - 2x \rfloor = 1$:

$1 \leq 5 - 2x < 2$ Solving for $x$: $1 \leq 5 - 2x \implies x \geq \frac{3}{2}$ $5 - 2x < 2 \implies x < 2$ Thus, for $\frac{3}{2} \leq x < 2$, $f(x) = 1$.

5. When $5 - 2x \in [0, 1)$, then $\lfloor 5 - 2x \rfloor = 0$:

$0 \leq 5 - 2x < 1$ Solving for $x$: $0 \leq 5 - 2x \implies x \geq 2$ $5 - 2x < 1 \implies x < \frac{5}{2}$ Thus, for $2 \leq x < \frac{5}{2}$, $f(x) = 0$.

This process can be continued for negative integers, covering the entire real line.

The piecewise function for $f(x) = \lfloor 5 - 2x \rfloor$ is therefore:

\begin{cases} 4, & 0 < x \leq \frac{1}{2} \\ 3, & \frac{1}{2} < x \leq 1 \\ 2, & 1 < x \leq \frac{3}{2} \\ 1, & \frac{3}{2} < x \leq 2 \\ 0, & 2 < x \leq 3 \end{cases}$$ Would you like further clarification or details on a specific part? Here are 5 related questions you might find interesting: 1. How does the behavior of a floor function differ from a ceiling function? 2. Can a piecewise function involve non-integer values for floor functions? 3. How would the piecewise function change if it were $$\lceil 5 - 2x \rceil$$ instead? 4. How can we express the floor function for more complex functions like $$\lfloor x^2 - 3x \rfloor$$? 5. How does the floor function behave as $$x$$ approaches infinity? **Tip**: The floor function can be a useful tool for simplifying expressions where rounding down to the nearest integer is required.