The maximum value of `(log x)/( x)` is

Options

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

Answer

To find the maximum value of the function \( f(x) = \frac{\log x}{x} \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) using the quotient rule. The quotient rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:

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

For our function, let:
- \( u = \log x \) (thus \( \frac{du}{dx} = \frac{1}{x} \))
- \( v = x \) (thus \( \frac{dv}{dx} = 1 \))

Applying 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 to zero
To find the critical points, we set the derivative equal to zero:

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

This implies:

\[ 1 - \log x = 0 \quad \Rightarrow \quad \log x = 1 \quad \Rightarrow \quad x = e \]

### Step 3: Determine if it's a maximum or minimum
Next, we need to check the second derivative to determine if this critical point is a maximum or minimum. We will differentiate \( f'(x) \):

Using the quotient rule again on \( f'(x) = \frac{1 - \log x}{x^2} \):

Let:
- \( u = 1 - \log x \) (thus \( \frac{du}{dx} = -\frac{1}{x} \))
- \( v = x^2 \) (thus \( \frac{dv}{dx} = 2x \))

Applying the quotient rule:

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

Simplifying this gives:

\[ f''(x) = \frac{-x - 2x(1 - \log x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2x \log x - 3x}{x^4} = \frac{2 \log x - 3}{x^3} \]

### Step 4: Evaluate the second derivative at \( x = e \)
Now we evaluate \( f''(e) \):

\[ f''(e) = \frac{2 \log e - 3}{e^3} = \frac{2 \cdot 1 - 3}{e^3} = \frac{2 - 3}{e^3} = \frac{-1}{e^3} \]

Since \( f''(e) < 0 \), this indicates that \( x = e \) is a local maximum.

### Step 5: Find the maximum value
Now we find the maximum value of \( f(x) \) at \( x = e \):

\[ f(e) = \frac{\log e}{e} = \frac{1}{e} \]

Thus, the maximum value of \( \frac{\log x}{x} \) is \( \frac{1}{e} \).

### Final Answer
The maximum value of \( \frac{\log x}{x} \) is \( \frac{1}{e} \).

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