Piecewise Functions Worksheet with evaluation and graphing exercises.

Worksheet - Piecewise Functions

#### Problem 1: Evaluate the following for \( f(x) = \begin{cases}
3x - 5, & x > 4 \\
x^2, & x \leq 4
\end{cases} \)

1. Evaluate \( f(7) \):
- Since \( 7 > 4 \), we use the first piece of the function: \( f(x) = 3x - 5 \).
- Substitute \( x = 7 \):
\[
f(7) = 3(7) - 5 = 21 - 5 = 16
\]
- Answer: \( f(7) = 16 \)

2. Evaluate \( f(4) \):
- Since \( 4 \leq 4 \), we use the second piece of the function: \( f(x) = x^2 \).
- Substitute \( x = 4 \):
\[
f(4) = 4^2 = 16
\]
- Answer: \( f(4) = 16 \)

3. Evaluate \( f(-3) \):
- Since \( -3 \leq 4 \), we use the second piece of the function: \( f(x) = x^2 \).
- Substitute \( x = -3 \):
\[
f(-3) = (-3)^2 = 9
\]
- Answer: \( f(-3) = 9 \)

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#### Problem 2: Evaluate the following for \( f(x) = \begin{cases}
-2|x + 1|, & x \leq 1 \\
3, & 1 < x < 3 \\
6 - 2x, & x \geq 3
\end{cases} \)

4. Evaluate \( f(10) \):
- Since \( 10 \geq 3 \), we use the third piece of the function: \( f(x) = 6 - 2x \).
- Substitute \( x = 10 \):
\[
f(10) = 6 - 2(10) = 6 - 20 = -14
\]
- Answer: \( f(10) = -14 \)

5. Evaluate \( f(2) \):
- Since \( 1 < 2 < 3 \), we use the second piece of the function: \( f(x) = 3 \).
- Answer: \( f(2) = 3 \)

6. Evaluate \( f(0) \):
- Since \( 0 \leq 1 \), we use the first piece of the function: \( f(x) = -2|x + 1| \).
- Substitute \( x = 0 \):
\[
f(0) = -2|0 + 1| = -2|1| = -2(1) = -2
\]
- Answer: \( f(0) = -2 \)

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#### Problem 3: Graph the following piecewise functions.

7. **Graph \( f(x) = \begin{cases}
-2, & x < 0 \\
3, & x \geq 0
\end{cases} \)**

- For \( x < 0 \), the function is a constant \( f(x) = -2 \). This is a horizontal line at \( y = -2 \) extending to the left.
- For \( x \geq 0 \), the function is a constant \( f(x) = 3 \). This is a horizontal line at \( y = 3 \) extending to the right.
- At \( x = 0 \), the function jumps from \( y = -2 \) to \( y = 3 \). The graph should show an open circle at \( (0, -2) \) and a closed circle at \( (0, 3) \).

8. **Graph \( g(x) = \begin{cases}
-x + 2, & x < 2 \\
x - 2, & x \geq 2
\end{cases} \)**

- For \( x < 2 \), the function is a linear equation \( g(x) = -x + 2 \). This is a straight line with a slope of \(-1\) and a y-intercept of \(2\).
- For \( x \geq 2 \), the function is a linear equation \( g(x) = x - 2 \). This is a straight line with a slope of \(1\) and a y-intercept of \(-2\).
- At \( x = 2 \), the function changes from \( g(x) = -x + 2 \) to \( g(x) = x - 2 \). The value at \( x = 2 \) is \( g(2) = 2 - 2 = 0 \). The graph should show an open circle at the end of the first line and a closed circle at the start of the second line.

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Final Answers:

1. \( f(7) = 16 \)
2. \( f(4) = 16 \)
3. \( f(-3) = 9 \)
4. \( f(10) = -14 \)
5. \( f(2) = 3 \)
6. \( f(0) = -2 \)

\[
\boxed{16, 16, 9, -14, 3, -2}
\]