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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 :

GinnyAnswer
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]
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