Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Rewrite the given function for better readability and correct any errors.

[tex]\[
f(x) =
\begin{cases}
\log_2(x+4), & x \leq 0 \\
x + 3, & 0 \ \textless \ x \ \textless \ 2 \\
x^2 - 6x + 2, & x \geq 3
\end{cases}
\][/tex]


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]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.