The function f(x) = `(x)/(log x)` increases on the interval

Options

A. `(0,oo)`
B. `(0,e)`
C. `(e,oo)`
D. none of these

Answer

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

### Step 1: Find the derivative of the function
We will use the quotient rule to differentiate \( f(x) \). The quotient rule states that if \( f(x) = \frac{u}{v} \), then:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
For our function, let:
- \( u = x \) and \( v = \log x \)
- Therefore, \( u' = 1 \) and \( v' = \frac{1}{x} \)

Now applying the quotient rule:
\[ f'(x) = \frac{(1)(\log x) - (x)\left(\frac{1}{x}\right)}{(\log x)^2} \]
This simplifies to:
\[ f'(x) = \frac{\log x - 1}{(\log x)^2} \]

### Step 2: Set the derivative greater than zero
To find where the function is increasing, we set the derivative greater than zero:
\[ \frac{\log x - 1}{(\log x)^2} > 0 \]
The denominator \( (\log x)^2 \) is always positive for \( x > 1 \) (since \( \log x \) is defined and positive). Therefore, we only need to consider the numerator:
\[ \log x - 1 > 0 \]
This simplifies to:
\[ \log x > 1 \]

### Step 3: Solve the inequality
Exponentiating both sides gives:
\[ x > e \]
Thus, the function \( f(x) \) is increasing when \( x \) is greater than \( e \).

### Step 4: State the interval
The interval on which the function is increasing is:
\[ (x \in (e, \infty)) \]

### Final Answer
The function \( f(x) = \frac{x}{\log x} \) increases on the interval \( (e, \infty) \).

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