Given that `f(x) = x^(1//x), x gt 0` has the maximum value at x = e, then

Options

A. `e^(pi) gt pi^e`
B. `e^pi lt pi^e`
C. `e^pi = pi^e`
D. `e^pi le pi^e`

Answer

To solve the problem, we need to analyze the function \( f(x) = x^{1/x} \) and confirm that it achieves its maximum value at \( x = e \). We will also derive the inequality \( e^\pi > \pi^e \).

### Step-by-Step Solution:

1. **Define the function**:
We start with the function:
\[ f(x) = x^{1/x} \]
where \( x > 0 \).

2. **Find the maximum point**:
We know that the maximum value occurs at \( x = e \). To confirm this, we can take the natural logarithm of \( f(x) \):
\[ \ln(f(x)) = \frac{1}{x} \ln(x) \]

3. **Differentiate the function**:
We differentiate \( \ln(f(x)) \) using the product and chain rules:
\[ \frac{d}{dx} \ln(f(x)) = \frac{d}{dx} \left( \frac{\ln(x)}{x} \right) \]
Using the quotient rule:
\[ \frac{d}{dx} \left( \frac{\ln(x)}{x} \right) = \frac{x \cdot \frac{1}{x} - \ln(x)}{x^2} = \frac{1 - \ln(x)}{x^2} \]

4. **Set the derivative to zero**:
To find critical points, set the derivative equal to zero:
\[ 1 - \ln(x) = 0 \implies \ln(x) = 1 \implies x = e \]

5. **Determine the nature of the critical point**:
To confirm that this point is a maximum, we can check the second derivative or observe the behavior of the first derivative around \( x = e \):
- For \( x < e \), \( \ln(x) < 1 \) so \( 1 - \ln(x) > 0 \) (increasing).
- For \( x > e \), \( \ln(x) > 1 \) so \( 1 - \ln(x) < 0 \) (decreasing).
This confirms that \( x = e \) is indeed a maximum.

6. **Compare values at specific points**:
We need to show that \( f(\pi) < f(e) \). This translates to:
\[ \pi^{1/\pi} < e^{1/e} \]

7. **Raise both sides to the power of \( e \cdot \pi \)**:
To eliminate the fractional exponents, we raise both sides to \( e \cdot \pi \):
\[ \left( \pi^{1/\pi} \right)^{e \cdot \pi} < \left( e^{1/e} \right)^{e \cdot \pi} \]
Simplifying gives:
\[ \pi^e < e^\pi \]

8. **Conclusion**:
Thus, we have shown that:
\[ e^\pi > \pi^e \]