The product of minimum value of `x^(x)` and maximum value of `((1)/(x))^(x)` is

Options

A. `e`
B. `e^(-1)`
C. `1`
D. `e^(2)`

Answer

To solve the problem, we need to find the product of the minimum value of \( x^x \) and the maximum value of \( \left(\frac{1}{x}\right)^x \).

### Step 1: Finding the Minimum Value of \( x^x \)

1. **Define the function:** Let \( y_1 = x^x \).
2. **Take the natural logarithm:**
\[ \log y_1 = x \log x \]
3. **Differentiate using the product rule:**
\[ \frac{1}{y_1} \frac{dy_1}{dx} = \log x + 1 \]
Therefore,
\[ \frac{dy_1}{dx} = y_1 (\log x + 1) = x^x (\log x + 1) \]
4. **Set the derivative to zero for critical points:**
\[ x^x (\log x + 1) = 0 \]
Since \( x^x \) is always positive for \( x > 0 \), we set:
\[ \log x + 1 = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e} \]
5. **Find the minimum value:**
\[ y_1 \text{ at } x = \frac{1}{e} \implies y_1 = \left(\frac{1}{e}\right)^{\frac{1}{e}} = \frac{1}{e^{1/e}} \]

### Step 2: Finding the Maximum Value of \( \left(\frac{1}{x}\right)^x \)

1. **Define the function:** Let \( y_2 = \left(\frac{1}{x}\right)^x = x^{-x} \).
2. **Take the natural logarithm:**
\[ \log y_2 = -x \log x \]
3. **Differentiate using the product rule:**
\[ \frac{1}{y_2} \frac{dy_2}{dx} = -\log x - 1 \]
Therefore,
\[ \frac{dy_2}{dx} = y_2 (-\log x - 1) = x^{-x} (-\log x - 1) \]
4. **Set the derivative to zero for critical points:**
\[ x^{-x} (-\log x - 1) = 0 \]
Again, since \( x^{-x} \) is always positive for \( x > 0 \), we set:
\[ -\log x - 1 = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e} \]
5. **Find the maximum value:**
\[ y_2 \text{ at } x = \frac{1}{e} \implies y_2 = \left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}} = e^{1/e} \]

### Step 3: Calculate the Product

Now, we need to find the product of the minimum value of \( y_1 \) and the maximum value of \( y_2 \):
\[ \text{Product} = \left(\frac{1}{e^{1/e}}\right) \cdot \left(e^{1/e}\right) = \frac{1}{e^{1/e}} \cdot e^{1/e} = 1 \]

### Final Answer

Thus, the product of the minimum value of \( x^x \) and the maximum value of \( \left(\frac{1}{x}\right)^x \) is \( \boxed{1} \).