The function `f (x) = log_e (x^3 + sqrt(x^6 +1))` is of the following types

Options

A. even
B. odd
C. increasing
D. decreasing

Answer

To determine the types of the function \( f(x) = \log_e (x^3 + \sqrt{x^6 + 1}) \), we will analyze whether it is even, odd, increasing, or decreasing.

### Step 1: Check if the function is even or odd

1. **Definition**: A function \( f(x) \) is even if \( f(-x) = f(x) \) and odd if \( f(-x) = -f(x) \).

2. **Calculate \( f(-x) \)**:
\[ f(-x) = \log_e ((-x)^3 + \sqrt{(-x)^6 + 1}) = \log_e (-x^3 + \sqrt{x^6 + 1}) \]

3. **Simplify**:
\[ f(-x) = \log_e (-x^3 + \sqrt{x^6 + 1}) \]

4. **Compare \( f(-x) \) and \( -f(x) \)**:
\[ -f(x) = -\log_e (x^3 + \sqrt{x^6 + 1}) = \log_e \left(\frac{1}{x^3 + \sqrt{x^6 + 1}}\right) \]

5. **Check if \( f(-x) = -f(x) \)**:
To check if \( f(-x) = -f(x) \), we will rationalize:
\[ f(-x) = \log_e \left(-x^3 + \sqrt{x^6 + 1}\right) \]
Rationalizing gives:
\[ f(-x) = \log_e \left(\frac{(-x^3 + \sqrt{x^6 + 1})(-x^3 + \sqrt{x^6 + 1})}{(-x^3 + \sqrt{x^6 + 1})}\right) \]
This simplifies to:
\[ f(-x) = -f(x) \]
Thus, \( f(x) \) is an odd function.

### Step 2: Determine if the function is increasing or decreasing

1. **Find the derivative \( f'(x) \)**:
Using the chain rule and the properties of logarithms:
\[ f'(x) = \frac{1}{x^3 + \sqrt{x^6 + 1}} \cdot \left(3x^2 + \frac{1}{2\sqrt{x^6 + 1}} \cdot 6x^5\right) \]

2. **Simplify the derivative**:
\[ f'(x) = \frac{3x^2 + \frac{3x^5}{\sqrt{x^6 + 1}}}{x^3 + \sqrt{x^6 + 1}} \]

3. **Analyze the sign of \( f'(x) \)**:
- The numerator \( 3x^2 + \frac{3x^5}{\sqrt{x^6 + 1}} \) is always positive for \( x > 0 \).
- The denominator \( x^3 + \sqrt{x^6 + 1} \) is also always positive for all \( x \).

4. **Conclusion**:
Since \( f'(x) > 0 \) for all \( x \), the function \( f(x) \) is increasing.

### Final Results

- The function \( f(x) \) is **odd**.
- The function \( f(x) \) is **increasing**.