If `alpha lt 0`, the function `(e^( alpha x) +e^(- alpha x ))` is a monotonic decreasing function for all values of x, where

Options

A. ` x gt 0`
B. ` x lt 0`
C. ` x gt 1`
D. ` x lt 1`

Answer

To determine whether the function \( f(x) = e^{\alpha x} + e^{-\alpha x} \) is a monotonic decreasing function for all values of \( x \) when \( \alpha < 0 \), we will follow these steps:

### Step 1: Differentiate the function
To analyze the monotonicity of the function, we first need to find its derivative.

\[ f'(x) = \frac{d}{dx}(e^{\alpha x} + e^{-\alpha x}) \]

Using the chain rule, we differentiate each term:

\[ f'(x) = \alpha e^{\alpha x} - \alpha e^{-\alpha x} \]

### Step 2: Factor out common terms
We can factor out \( \alpha \) from the derivative:

\[ f'(x) = \alpha \left( e^{\alpha x} - e^{-\alpha x} \right) \]

### Step 3: Analyze the sign of the derivative
Since \( \alpha < 0 \), we need to analyze the term \( e^{\alpha x} - e^{-\alpha x} \).

1. **For \( x < 0 \)**:
- \( e^{\alpha x} \) is a positive number less than 1 (since \( \alpha < 0 \)).
- \( e^{-\alpha x} \) is a positive number greater than 1.
- Therefore, \( e^{\alpha x} - e^{-\alpha x} < 0 \).

So, \( f'(x) < 0 \) when \( x < 0 \).

2. **For \( x = 0 \)**:
- \( e^{\alpha \cdot 0} - e^{-\alpha \cdot 0} = 1 - 1 = 0 \).

3. **For \( x > 0 \)**:
- \( e^{\alpha x} \) is a positive number less than 1.
- \( e^{-\alpha x} \) is a positive number greater than 1.
- Therefore, \( e^{\alpha x} - e^{-\alpha x} < 0 \).

Thus, \( f'(x) < 0 \) for all \( x \neq 0 \).

### Conclusion
Since \( f'(x) < 0 \) for all \( x < 0 \) and \( x > 0 \), the function \( f(x) = e^{\alpha x} + e^{-\alpha x} \) is a monotonic decreasing function for all values of \( x \) when \( \alpha < 0 \).