Evaluating Piecewise Functions online exercise for - Free Printable

Problem: Evaluating Piecewise Functions

We are tasked with evaluating piecewise functions for specific values of \( x \). Let's solve each problem step by step.

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Problem 1:

The function is:
\[
f(x) =
\begin{cases}
-x^2 - 1, & x \leq 2 \\
\frac{4}{5}x - 4, & x > 2
\end{cases}
\]

#### a. \( f(0) \)
- Since \( 0 \leq 2 \), we use the first piece of the function: \( f(x) = -x^2 - 1 \).
- Substitute \( x = 0 \):
\[
f(0) = -(0)^2 - 1 = -1
\]
- Answer: \( f(0) = -1 \)

#### b. \( f(5) \)
- Since \( 5 > 2 \), we use the second piece of the function: \( f(x) = \frac{4}{5}x - 4 \).
- Substitute \( x = 5 \):
\[
f(5) = \frac{4}{5}(5) - 4 = 4 - 4 = 0
\]
- Answer: \( f(5) = 0 \)

#### c. \( f(2) \)
- Since \( 2 \leq 2 \), we use the first piece of the function: \( f(x) = -x^2 - 1 \).
- Substitute \( x = 2 \):
\[
f(2) = -(2)^2 - 1 = -4 - 1 = -5
\]
- Answer: \( f(2) = -5 \)

#### d. \( f(-3) \)
- Since \( -3 \leq 2 \), we use the first piece of the function: \( f(x) = -x^2 - 1 \).
- Substitute \( x = -3 \):
\[
f(-3) = -(-3)^2 - 1 = -9 - 1 = -10
\]
- Answer: \( f(-3) = -10 \)

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Problem 2:

The function is:
\[
f(x) =
\begin{cases}
x^3 - 7x, & x \leq -3 \\
8, & -3 < x \leq 3 \\
\sqrt{2x + 3}, & x > 3
\end{cases}
\]

#### a. \( f(-5) \)
- Since \( -5 \leq -3 \), we use the first piece of the function: \( f(x) = x^3 - 7x \).
- Substitute \( x = -5 \):
\[
f(-5) = (-5)^3 - 7(-5) = -125 + 35 = -90
\]
- Answer: \( f(-5) = -90 \)

#### b. \( f(11) \)
- Since \( 11 > 3 \), we use the third piece of the function: \( f(x) = \sqrt{2x + 3} \).
- Substitute \( x = 11 \):
\[
f(11) = \sqrt{2(11) + 3} = \sqrt{22 + 3} = \sqrt{25} = 5
\]
- Answer: \( f(11) = 5 \)

#### c. \( f(0) \)
- Since \( -3 < 0 \leq 3 \), we use the second piece of the function: \( f(x) = 8 \).
- Answer: \( f(0) = 8 \)

#### d. \( f(3) \)
- Since \( -3 < 3 \leq 3 \), we use the second piece of the function: \( f(x) = 8 \).
- Answer: \( f(3) = 8 \)

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Problem 3:

The function is:
\[
f(x) =
\begin{cases}
\frac{3}{x+4}, & x < -5 \\
x^2 - 3x, & -5 < x \leq 0 \\
x^4 - 7, & x > 0
\end{cases}
\]

#### a. \( f(-1) \)
- Since \( -5 < -1 \leq 0 \), we use the second piece of the function: \( f(x) = x^2 - 3x \).
- Substitute \( x = -1 \):
\[
f(-1) = (-1)^2 - 3(-1) = 1 + 3 = 4
\]
- Answer: \( f(-1) = 4 \)

#### b. \( f(4) \)
- Since \( 4 > 0 \), we use the third piece of the function: \( f(x) = x^4 - 7 \).
- Substitute \( x = 4 \):
\[
f(4) = (4)^4 - 7 = 256 - 7 = 249
\]
- Answer: \( f(4) = 249 \)

#### c. \( f(-10) \)
- Since \( -10 < -5 \), we use the first piece of the function: \( f(x) = \frac{3}{x+4} \).
- Substitute \( x = -10 \):
\[
f(-10) = \frac{3}{-10 + 4} = \frac{3}{-6} = -\frac{1}{2}
\]
- Answer: \( f(-10) = -\frac{1}{2} \)

#### d. \( f(0) \)
- Since \( -5 < 0 \leq 0 \), we use the second piece of the function: \( f(x) = x^2 - 3x \).
- Substitute \( x = 0 \):
\[
f(0) = (0)^2 - 3(0) = 0
\]
- Answer: \( f(0) = 0 \)

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Problem 4:

The function is:
\[
f(x) =
\begin{cases}
|2x + 7|, & x \leq -4 \\
1 + x^2, & -4 < x \leq 1 \\
6, & 1 < x < 3 \\
\frac{1}{3}x + 8, & x \geq 3
\end{cases}
\]

#### a. \( f(5) \)
- Since \( 5 \geq 3 \), we use the fourth piece of the function: \( f(x) = \frac{1}{3}x + 8 \).
- Substitute \( x = 5 \):
\[
f(5) = \frac{1}{3}(5) + 8 = \frac{5}{3} + 8 = \frac{5}{3} + \frac{24}{3} = \frac{29}{3}
\]
- Answer: \( f(5) = \frac{29}{3} \)

#### b. \( f(1) \)
- Since \( -4 < 1 \leq 1 \), we use the second piece of the function: \( f(x) = 1 + x^2 \).
- Substitute \( x = 1 \):
\[
f(1) = 1 + (1)^2 = 1 + 1 = 2
\]
- Answer: \( f(1) = 2 \)

#### c. \( f(-4) \)
- Since \( -4 \leq -4 \), we use the first piece of the function: \( f(x) = |2x + 7| \).
- Substitute \( x = -4 \):
\[
f(-4) = |2(-4) + 7| = |-8 + 7| = |-1| = 1
\]
- Answer: \( f(-4) = 1 \)

#### d. \( f(2) \)
- Since \( 1 < 2 < 3 \), we use the third piece of the function: \( f(x) = 6 \).
- Answer: \( f(2) = 6 \)

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Problem 5:

The graph shows a piecewise function with different segments. We need to evaluate \( f(x) \) at specific points.

#### a. \( f(-1) \)
- From the graph, when \( x = -1 \), the corresponding \( y \)-value is \( 0 \).
- Answer: \( f(-1) = 0 \)

#### b. \( f(2) \)
- From the graph, when \( x = 2 \), the corresponding \( y \)-value is \( 2 \).
- Answer: \( f(2) = 2 \)

#### c. \( f(1) \)
- From the graph, when \( x = 1 \), the corresponding \( y \)-value is \( 1 \).
- Answer: \( f(1) = 1 \)

#### d. \( f(-2) \)
- From the graph, when \( x = -2 \), the corresponding \( y \)-value is \( -1 \).
- Answer: \( f(-2) = -1 \)

#### e. \( f(0) \)
- From the graph, when \( x = 0 \), the corresponding \( y \)-value is \( 0 \).
- Answer: \( f(0) = 0 \)

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Problem 6:

The graph shows another piecewise function with different segments. We need to evaluate \( f(x) \) at specific points.

#### a. \( f(-3) \)
- From the graph, when \( x = -3 \), the corresponding \( y \)-value is \( -2 \).
- Answer: \( f(-3) = -2 \)

#### b. \( f(4) \)
- From the graph, when \( x = 4 \), the corresponding \( y \)-value is \( 3 \).
- Answer: \( f(4) = 3 \)

#### c. \( f(1) \)
- From the graph, when \( x = 1 \), the corresponding \( y \)-value is \( 2 \).
- Answer: \( f(1) = 2 \)

#### d. \( f(-1) \)
- From the graph, when \( x = -1 \), the corresponding \( y \)-value is \( 0 \).
- Answer: \( f(-1) = 0 \)

#### e. \( f(0) \)
- From the graph, when \( x = 0 \), the corresponding \( y \)-value is \( -1 \).
- Answer: \( f(0) = -1 \)

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Final Answers:

1.
- \( f(0) = -1 \)
- \( f(5) = 0 \)
- \( f(2) = -5 \)
- \( f(-3) = -10 \)

2.
- \( f(-5) = -90 \)
- \( f(11) = 5 \)
- \( f(0) = 8 \)
- \( f(3) = 8 \)

3.
- \( f(-1) = 4 \)
- \( f(4) = 249 \)
- \( f(-10) = -\frac{1}{2} \)
- \( f(0) = 0 \)

4.
- \( f(5) = \frac{29}{3} \)
- \( f(1) = 2 \)
- \( f(-4) = 1 \)
- \( f(2) = 6 \)

5.
- \( f(-1) = 0 \)
- \( f(2) = 2 \)
- \( f(1) = 1 \)
- \( f(-2) = -1 \)
- \( f(0) = 0 \)

6.
- \( f(-3) = -2 \)
- \( f(4) = 3 \)
- \( f(1) = 2 \)
- \( f(-1) = 0 \)
- \( f(0) = -1 \)

\boxed{\text{See detailed solutions above}}