Let the function `f(x)=3x^(2)-4x+8log(1+|x|)` be defined on the interval [0,1]. The even extension of f(x) to the interval [0,1]. The even extension of f(x) to the interval [-1,1] is

Options

A. `3x^(2)+4x+8 log(1+|x|)`
B. `3x^(2)-4x+8 log (1+|x|)`
C. `3x^(2)+4x-8 log(1+|x|)`
D. none of these

Answer

To find the even extension of the function \( f(x) = 3x^2 - 4x + 8\log(1 + |x|) \) defined on the interval \([0, 1]\) to the interval \([-1, 1]\), we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Even Extension**:
The even extension of a function \( f(x) \) is defined as:
\[ f(-x) = f(x) \quad \text{for } x \in [0, 1] \]
This means that for every \( x \) in the interval \([0, 1]\), the value of the function at \(-x\) will be equal to the value of the function at \( x \).

2. **Calculate \( f(-x) \)**:
We need to substitute \(-x\) into the function \( f(x) \):
\[ f(-x) = 3(-x)^2 - 4(-x) + 8\log(1 + |-x|) \]
Simplifying this:
\[ f(-x) = 3x^2 + 4x + 8\log(1 + |x|) \]

3. **Define the Even Extension**:
The even extension of \( f(x) \) on the interval \([-1, 1]\) can be defined as:
\[ f(x) = \begin{cases} 3x^2 - 4x + 8\log(1 + |x|) & \text{for } x \in [0, 1] \\ 3x^2 + 4x + 8\log(1 + |x|) & \text{for } x \in [-1, 0] \end{cases} \]

4. **Final Formulation**:
Thus, the even extension of \( f(x) \) to the interval \([-1, 1]\) is:
\[ f(x) = \begin{cases} 3x^2 - 4x + 8\log(1 + |x|) & \text{for } x \in [0, 1] \\ 3x^2 + 4x + 8\log(1 + |x|) & \text{for } x \in [-1, 0] \end{cases} \]