Write f(x) = [5−2x] as piecewise function. Note that [5 - 2x] is a floor function 


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:

\[
f(x) =
\begin{cases}
4, & 0 < x \leq \frac{1}{2} \\
3, & \frac{1}{2} < x \leq 1 \\
2, & 1 < x \leq \frac{3}{2} \\
1, & \frac{3}{2} < x \leq 2 \\
0, & 2 < x \leq 3
\end{cases}
\]

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.

Write f(x) = [5−2x] as piecewise function. Note that [5 - 2x] is a floor function.

Discover how to express f(x) = ⌊5 - 2x⌋ as a piecewise function by breaking it down into intervals where the floor function outputs a constant integer. This step-by-step guide explains the use of inequalities and interval notation. Ideal for high school students in Grades 10-12.

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