The function `f(x)=x log x`

Options

A. has maximum value `(1)/(e )`
B. has maximum value `(-1)/(e )`
C. has minimum value `(1)/(e )`
D. has minimum value `(-1)/(e )`

Answer

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

### Step 1: Differentiate the function
We start by finding the first derivative of the function. We will use the product rule for differentiation, which states that if \( u = x \) and \( v = \log x \), then:

\[ f'(x) = u'v + uv' \]

Here, \( u' = 1 \) and \( v' = \frac{1}{x} \). Thus,

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

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

\[ \log x + 1 = 0 \]

This simplifies to:

\[ \log x = -1 \]

### Step 3: Solve for \( x \)
Exponentiating both sides gives us:

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

### Step 4: Determine if it is a maximum or minimum
Next, we need to check the second derivative to determine if this critical point is a maximum or minimum. We differentiate \( f'(x) \):

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

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

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

Since \( f''\left(\frac{1}{e}\right) > 0 \), this indicates that \( x = \frac{1}{e} \) is a local minimum.

### Step 6: Find the minimum value
Finally, we substitute \( x = \frac{1}{e} \) back into the original function to find the minimum value:

\[ f\left(\frac{1}{e}\right) = \frac{1}{e} \log\left(\frac{1}{e}\right) = \frac{1}{e} \cdot (-1) = -\frac{1}{e} \]

### Conclusion
Thus, the minimum value of the function \( f(x) = x \log x \) occurs at \( x = \frac{1}{e} \) and is \( -\frac{1}{e} \).

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