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

Options

A. `(1,2e)`
B. `(0,e)`
C. `(2,2e)`
D. `(1//e ,2e)`

Answer

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

### Step 1: Find the derivative of the function
To find the intervals where the function is increasing, we first need to calculate the derivative \( f'(x) \).

Using the quotient rule for differentiation, where \( u = \log x \) and \( v = x \):

\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]

Here,
- \( u' = \frac{1}{x} \) (the derivative of \( \log x \))
- \( v' = 1 \) (the derivative of \( x \))

Substituting these into the quotient rule:

\[ f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2} \]

### Step 2: Set the derivative greater than zero
To find where the function is increasing, we need to solve the inequality:

\[ f'(x) > 0 \]

This leads to:

\[ \frac{1 - \log x}{x^2} > 0 \]

Since \( x^2 > 0 \) for all \( x > 0 \), we only need to consider the numerator:

\[ 1 - \log x > 0 \]

### Step 3: Solve the inequality
Rearranging the inequality gives:

\[ \log x < 1 \]

Exponentiating both sides (with base \( e \)) results in:

\[ x < e \]

### Step 4: Consider the domain of the function
The logarithmic function \( \log x \) is defined for \( x > 0 \). Therefore, we combine this with our previous result:

\[ 0 < x < e \]

### Conclusion
The function \( f(x) = \frac{\log x}{x} \) is increasing in the interval \( (0, e) \).

### Final Answer
The function \( f(x) \) is increasing in the interval \( (0, e) \).

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