Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Piecewise Function and Evaluations
Suppose that the function \( f \) is defined on the interval \((-2.5, 1.5)\) as follows:
\[
f(x) =
\begin{cases}
-2 & \text{if } -2.5 < x \leq -1.5 \\
-1 & \text{if } -1.5 < x \leq -0.5 \\
0 & \text{if } -0.5 < x < 0.5 \\
1 & \text{if } 0.5 \leq x < 1.5
\end{cases}
\]
We are given a function \( f \) defined over different intervals of \( x \), and we need to determine the values of the function at specific points.
#### Task
Find \( f(-1.5) \), \( f(-0.7) \), and \( f(0.5) \).
#### Solution
For each specific value of \( x \), we will refer to the conditions of the piecewise function to determine \( f(x) \):
1. \( f(-1.5) \)
According to the piecewise definition, for \( -1.5 < x \leq -0.5 \), \( f(x) = -1 \). Thus:
\[
f(-1.5) = -1
\]
2. \( f(-0.7) \)
For \( -0.5 < x < 0.5 \), \( f(x) = 0 \). Thus:
\[
f(-0.7) = 0
\]
3. \( f(0.5) \)
For \( 0.5 \leq x < 1.5 \), \( f(x) = 1 \). Thus:
\[
f(0.5) = 1
\]
Substituting these values into the evaluations:
\[
\begin{align*}
f(-1.5) &= -1 \\
f(-0.7) &= 0 \\
f(0.5) &= 1
\end{align*}
\]
This piecewise function separates \( x \) into four different intervals, each with a corresponding rule for calculating \( f(x) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a663202-2562-419a-bf45-34f83e9296f9%2Fa8d47c9a-7a97-4a8f-8a10-2306fd443adf%2Fbaj3kn_processed.png&w=3840&q=75)
Transcribed Image Text:### Piecewise Function and Evaluations
Suppose that the function \( f \) is defined on the interval \((-2.5, 1.5)\) as follows:
\[
f(x) =
\begin{cases}
-2 & \text{if } -2.5 < x \leq -1.5 \\
-1 & \text{if } -1.5 < x \leq -0.5 \\
0 & \text{if } -0.5 < x < 0.5 \\
1 & \text{if } 0.5 \leq x < 1.5
\end{cases}
\]
We are given a function \( f \) defined over different intervals of \( x \), and we need to determine the values of the function at specific points.
#### Task
Find \( f(-1.5) \), \( f(-0.7) \), and \( f(0.5) \).
#### Solution
For each specific value of \( x \), we will refer to the conditions of the piecewise function to determine \( f(x) \):
1. \( f(-1.5) \)
According to the piecewise definition, for \( -1.5 < x \leq -0.5 \), \( f(x) = -1 \). Thus:
\[
f(-1.5) = -1
\]
2. \( f(-0.7) \)
For \( -0.5 < x < 0.5 \), \( f(x) = 0 \). Thus:
\[
f(-0.7) = 0
\]
3. \( f(0.5) \)
For \( 0.5 \leq x < 1.5 \), \( f(x) = 1 \). Thus:
\[
f(0.5) = 1
\]
Substituting these values into the evaluations:
\[
\begin{align*}
f(-1.5) &= -1 \\
f(-0.7) &= 0 \\
f(0.5) &= 1
\end{align*}
\]
This piecewise function separates \( x \) into four different intervals, each with a corresponding rule for calculating \( f(x) \).
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