The function f(x) = x log x is minimum at x =

Options

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

Answer

To find the minimum of the function \( f(x) = x \log x \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) with respect to \( x \).

\[ f'(x) = \frac{d}{dx}(x \log x) \]

Using the product rule, where \( u = x \) and \( v = \log x \), we have:

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

Calculating \( \frac{dv}{dx} \) and \( \frac{du}{dx} \):

\[ \frac{dv}{dx} = \frac{1}{x}, \quad \frac{du}{dx} = 1 \]

Thus,

\[ f'(x) = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

\[ 1 + \log x = 0 \]

### Step 3: Solve for \( x \)
Solving for \( x \):

\[ \log x = -1 \]

Exponentiating both sides gives:

\[ x = e^{-1} = \frac{1}{e} \]

### Step 4: Determine if it is a minimum or maximum
To determine whether this critical point is a minimum or maximum, we will find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(1 + \log x) = \frac{1}{x} \]

### Step 5: Evaluate the second derivative at \( x = \frac{1}{e} \)
Now we evaluate \( f''(x) \) at \( x = \frac{1}{e} \):

\[ f''\left(\frac{1}{e}\right) = \frac{1}{\frac{1}{e}} = e \]

Since \( e > 0 \), this indicates that the function is concave up at \( x = \frac{1}{e} \), confirming that it is a minimum.

### Conclusion
Thus, the function \( f(x) = x \log x \) has a minimum at:

\[ x = \frac{1}{e} \]

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