Minimum value of the function `f(x)= (1/x)^(1//x)` is:

Options

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

Answer

To find the minimum value of the function \( f(x) = \left(\frac{1}{x}\right)^{\frac{1}{x}} \), we will follow these steps:

### Step 1: Rewrite the function
Let \( y = f(x) = \left(\frac{1}{x}\right)^{\frac{1}{x}} \).

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \log y = \frac{1}{x} \log\left(\frac{1}{x}\right) = \frac{1}{x} (-\log x) = -\frac{\log x}{x} \]

### Step 3: Differentiate using implicit differentiation
Now, differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = -\frac{d}{dx}\left(\frac{\log x}{x}\right) \]
Using the quotient rule for differentiation:
\[ \frac{d}{dx}\left(\frac{\log x}{x}\right) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2} \]
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = -\frac{1 - \log x}{x^2} \]
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = -y \cdot \frac{1 - \log x}{x^2} \]

### Step 4: Set the derivative to zero
To find critical points, set \( \frac{dy}{dx} = 0 \):
\[ -y \cdot \frac{1 - \log x}{x^2} = 0 \]
Since \( y \neq 0 \), we have:
\[ 1 - \log x = 0 \implies \log x = 1 \implies x = e \]

### Step 5: Second derivative test
Now we need to check if this point is a minimum by calculating the second derivative. We already have:
\[ \frac{dy}{dx} = -y \cdot \frac{1 - \log x}{x^2} \]
We differentiate again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-y \cdot \frac{1 - \log x}{x^2}\right) \]
Using the product rule:
\[ \frac{d^2y}{dx^2} = -\left(\frac{dy}{dx} \cdot \frac{1 - \log x}{x^2} + y \cdot \frac{d}{dx}\left(\frac{1 - \log x}{x^2}\right)\right) \]
Calculate \( \frac{d}{dx}\left(\frac{1 - \log x}{x^2}\right) \) using the quotient rule:
\[ \frac{d}{dx}\left(\frac{1 - \log x}{x^2}\right) = \frac{x^2 \cdot (-\frac{1}{x}) - (1 - \log x) \cdot 2x}{x^4} = \frac{-x + 2(1 - \log x)}{x^3} \]

### Step 6: Evaluate the second derivative at \( x = e \)
Now substitute \( x = e \):
\[ \frac{d^2y}{dx^2} \text{ at } x = e \]
We find \( y = \left(\frac{1}{e}\right)^{\frac{1}{e}} = e^{-\frac{1}{e}} \).

Since the second derivative is positive at \( x = e \), we confirm that this is a minimum.

### Step 7: Minimum value
Thus, the minimum value of the function is:
\[ f(e) = \left(\frac{1}{e}\right)^{\frac{1}{e}} = e^{-\frac{1}{e}} \]

### Final Answer
The minimum value of the function \( f(x) = \left(\frac{1}{x}\right)^{\frac{1}{x}} \) is:
\[ \boxed{e^{-\frac{1}{e}}} \]