The function `f(x)=x^(1//x)` is increasing in the interval

Options

A. `(e,oo)`
B. `(-oo,e)`
C. `(-e,e)`
D. none of these

Answer

To determine the interval in which the function \( f(x) = x^{\frac{1}{x}} \) is increasing, we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = x^{\frac{1}{x}} \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we have:
\[ \log f(x) = \log \left( x^{\frac{1}{x}} \right) \]
Using the property of logarithms, this simplifies to:
\[ \log f(x) = \frac{1}{x} \log x \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\log f(x)) = \frac{d}{dx}\left(\frac{\log x}{x}\right) \]
Using the chain rule on the left side, we have:
\[ \frac{1}{f(x)} \frac{df}{dx} = \frac{(1)(x) - (\log x)(1)}{x^2} \]
This simplifies to:
\[ \frac{1}{f(x)} \frac{df}{dx} = \frac{1 - \log x}{x^2} \]

### Step 4: Solve for \( \frac{df}{dx} \)
Multiplying both sides by \( f(x) \):
\[ \frac{df}{dx} = f(x) \cdot \frac{1 - \log x}{x^2} \]

### Step 5: Determine when the function is increasing
The function \( f(x) \) is increasing when \( \frac{df}{dx} > 0 \). This occurs when:
\[ f(x) \cdot \frac{1 - \log x}{x^2} > 0 \]
Since \( f(x) \) and \( x^2 \) are always positive for \( x > 0 \), we need:
\[ 1 - \log x > 0 \]
This simplifies to:
\[ \log x < 1 \]
Exponentiating both sides gives:
\[ x < e \]

### Step 6: Conclusion
Thus, the function \( f(x) = x^{\frac{1}{x}} \) is increasing in the interval:
\[ (0, e) \]