Show that the maximum value of `(1/x)^x` is `e^(1//e)` .

Options

A. e
B. `e^( e)`
C. `e^(1//e)`
D. `(1/e)^(1//e)`

Explanation

To show that the maximum value of the function \( f(x) = \left(\frac{1}{x}\right)^x \) is \( e^{\frac{1}{e}} \), we will follow these steps:

### Step 1: Define the function
Let
\[ y = \left(\frac{1}{x}\right)^x \]
This can be rewritten as:
\[ y = x^{-x} \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we get:
\[ \ln y = \ln(x^{-x}) = -x \ln x \]

### Step 3: Differentiate using implicit differentiation
Now, differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \]
Using the product rule on the right side:
\[ \frac{d}{dx}(-x \ln x) = -\left(\ln x + 1\right) \]
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = -\left(\ln x + 1\right) \]

### Step 4: Solve for \(\frac{dy}{dx}\)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot -(\ln x + 1) \]
Substituting \( y = \left(\frac{1}{x}\right)^x \):
\[ \frac{dy}{dx} = \left(\frac{1}{x}\right)^x \cdot -(\ln x + 1) \]

### Step 5: Find critical points
To find the critical points, set \( \frac{dy}{dx} = 0 \):
\[ \left(\frac{1}{x}\right)^x \cdot -(\ln x + 1) = 0 \]
Since \( \left(\frac{1}{x}\right)^x \) is never zero, we have:
\[ -\ln x - 1 = 0 \implies \ln x = -1 \implies x = e^{-1} = \frac{1}{e} \]

### Step 6: Evaluate the function at the critical point
Now, substitute \( x = \frac{1}{e} \) back into the original function to find the maximum value:
\[ f\left(\frac{1}{e}\right) = \left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}} = e^{\frac{1}{e}} \]

### Conclusion
Thus, the maximum value of \( \left(\frac{1}{x}\right)^x \) is:
\[ \boxed{e^{\frac{1}{e}}} \]

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