The function f(x) =`e^(ax) +e^(-ax), agt 0` is monotonically increasing for

Options

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

Answer

To determine for which values of \( x \) the function \( f(x) = e^{ax} + e^{-ax} \) (where \( a > 0 \)) is monotonically increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(e^{ax}) + \frac{d}{dx}(e^{-ax}) \]
Using the chain rule, we get:
\[ f'(x) = ae^{ax} - ae^{-ax} \]

### Step 2: Set the derivative greater than zero
For the function to be monotonically increasing, we need:
\[ f'(x) > 0 \]
So, we set up the inequality:
\[ ae^{ax} - ae^{-ax} > 0 \]

### Step 3: Factor out \( a \)
Since \( a > 0 \), we can factor it out:
\[ a(e^{ax} - e^{-ax}) > 0 \]
This simplifies to:
\[ e^{ax} - e^{-ax} > 0 \]

### Step 4: Solve the inequality
We can rewrite the inequality:
\[ e^{ax} > e^{-ax} \]
Taking the natural logarithm of both sides (noting that the exponential function is always positive), we have:
\[ ax > -ax \]
This simplifies to:
\[ 2ax > 0 \]
Since \( a > 0 \), we can divide both sides by \( 2a \):
\[ x > 0 \]

### Conclusion
Thus, the function \( f(x) = e^{ax} + e^{-ax} \) is monotonically increasing for:
\[ x > 0 \]