Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Certainly! Let's go through the piecewise function [tex]\( f(x) \)[/tex] and find the value of [tex]\( f(x) \)[/tex] for different values of [tex]\( x \)[/tex].
The function is defined as:
[tex]\[ f(x) = \begin{cases} \log_2(x + 4) & x \leq 0 \\ x + 3 & 0 < x < 2 \\ x^2 - 6x + 2x & x \geq 3 \end{cases} \][/tex]
We'll evaluate the function for the specific values [tex]\( x = -5, -1, 0, 1, 3, 4 \)[/tex].
### 1. [tex]\( x = -5 \)[/tex]
Since [tex]\( -5 \leq 0 \)[/tex], we use the first piece of the function:
[tex]\[ f(-5) = \log_2(-5 + 4) = \log_2(-1) \][/tex]
However, the logarithm of a negative number is not defined in the real number system, so:
[tex]\[ f(-5) = \text{undefined} \][/tex]
### 2. [tex]\( x = -1 \)[/tex]
Since [tex]\( -1 \leq 0 \)[/tex], we use the first piece of the function:
[tex]\[ f(-1) = \log_2(-1 + 4) = \log_2(3) \approx 1.5849625 \][/tex]
So,
[tex]\[ f(-1) \approx 1.5849625 \][/tex]
### 3. [tex]\( x = 0 \)[/tex]
Since [tex]\( 0 \leq 0 \)[/tex], we use the first piece of the function:
[tex]\[ f(0) = \log_2(0 + 4) = \log_2(4) = 2 \][/tex]
So,
[tex]\[ f(0) = 2 \][/tex]
### 4. [tex]\( x = 1 \)[/tex]
Since [tex]\( 0 < 1 < 2 \)[/tex], we use the second piece of the function:
[tex]\[ f(1) = 1 + 3 = 4 \][/tex]
So,
[tex]\[ f(1) = 4 \][/tex]
### 5. [tex]\( x = 3 \)[/tex]
Since [tex]\( 3 \geq 3 \)[/tex], we use the third piece of the function:
[tex]\[ f(3) = 3^2 - 6 \cdot 3 + 2 \cdot 3 = 9 - 18 + 6 = -3 \][/tex]
So,
[tex]\[ f(3) = -3 \][/tex]
### 6. [tex]\( x = 4 \)[/tex]
Since [tex]\( 4 \geq 3 \)[/tex], we use the third piece of the function:
[tex]\[ f(4) = 4^2 - 6 \cdot 4 + 2 \cdot 4 = 16 - 24 + 8 = 0 \][/tex]
So,
[tex]\[ f(4) = 0 \][/tex]
### Summary
The function values for the given [tex]\( x \)[/tex] values are:
[tex]\[ f(-5) = \text{undefined} \][/tex]
[tex]\[ f(-1) \approx 1.5849625 \][/tex]
[tex]\[ f(0) = 2 \][/tex]
[tex]\[ f(1) = 4 \][/tex]
[tex]\[ f(3) = -3 \][/tex]
[tex]\[ f(4) = 0 \][/tex]
So, the computed results are:
[tex]\[ f(-5) = \text{NaN} \][/tex] (Not a Number due to undefined log)
[tex]\[ f(-1) \approx 1.5849625 \][/tex]
[tex]\[ f(0) = 2 \][/tex]
[tex]\[ f(1) = 4 \][/tex]
[tex]\[ f(3) = -3 \][/tex]
[tex]\[ f(4) = 0 \][/tex]
The function is defined as:
[tex]\[ f(x) = \begin{cases} \log_2(x + 4) & x \leq 0 \\ x + 3 & 0 < x < 2 \\ x^2 - 6x + 2x & x \geq 3 \end{cases} \][/tex]
We'll evaluate the function for the specific values [tex]\( x = -5, -1, 0, 1, 3, 4 \)[/tex].
### 1. [tex]\( x = -5 \)[/tex]
Since [tex]\( -5 \leq 0 \)[/tex], we use the first piece of the function:
[tex]\[ f(-5) = \log_2(-5 + 4) = \log_2(-1) \][/tex]
However, the logarithm of a negative number is not defined in the real number system, so:
[tex]\[ f(-5) = \text{undefined} \][/tex]
### 2. [tex]\( x = -1 \)[/tex]
Since [tex]\( -1 \leq 0 \)[/tex], we use the first piece of the function:
[tex]\[ f(-1) = \log_2(-1 + 4) = \log_2(3) \approx 1.5849625 \][/tex]
So,
[tex]\[ f(-1) \approx 1.5849625 \][/tex]
### 3. [tex]\( x = 0 \)[/tex]
Since [tex]\( 0 \leq 0 \)[/tex], we use the first piece of the function:
[tex]\[ f(0) = \log_2(0 + 4) = \log_2(4) = 2 \][/tex]
So,
[tex]\[ f(0) = 2 \][/tex]
### 4. [tex]\( x = 1 \)[/tex]
Since [tex]\( 0 < 1 < 2 \)[/tex], we use the second piece of the function:
[tex]\[ f(1) = 1 + 3 = 4 \][/tex]
So,
[tex]\[ f(1) = 4 \][/tex]
### 5. [tex]\( x = 3 \)[/tex]
Since [tex]\( 3 \geq 3 \)[/tex], we use the third piece of the function:
[tex]\[ f(3) = 3^2 - 6 \cdot 3 + 2 \cdot 3 = 9 - 18 + 6 = -3 \][/tex]
So,
[tex]\[ f(3) = -3 \][/tex]
### 6. [tex]\( x = 4 \)[/tex]
Since [tex]\( 4 \geq 3 \)[/tex], we use the third piece of the function:
[tex]\[ f(4) = 4^2 - 6 \cdot 4 + 2 \cdot 4 = 16 - 24 + 8 = 0 \][/tex]
So,
[tex]\[ f(4) = 0 \][/tex]
### Summary
The function values for the given [tex]\( x \)[/tex] values are:
[tex]\[ f(-5) = \text{undefined} \][/tex]
[tex]\[ f(-1) \approx 1.5849625 \][/tex]
[tex]\[ f(0) = 2 \][/tex]
[tex]\[ f(1) = 4 \][/tex]
[tex]\[ f(3) = -3 \][/tex]
[tex]\[ f(4) = 0 \][/tex]
So, the computed results are:
[tex]\[ f(-5) = \text{NaN} \][/tex] (Not a Number due to undefined log)
[tex]\[ f(-1) \approx 1.5849625 \][/tex]
[tex]\[ f(0) = 2 \][/tex]
[tex]\[ f(1) = 4 \][/tex]
[tex]\[ f(3) = -3 \][/tex]
[tex]\[ f(4) = 0 \][/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
