Difference between the greatest and the least values of the function `f(x)=x(ln x-2)` on `[1,e^2]` is

Options

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

Answer

To find the difference between the greatest and least values of the function \( f(x) = x(\ln x - 2) \) on the interval \([1, e^2]\), we will follow these steps:

### Step 1: Find the derivative of the function
We will use the product rule to differentiate \( f(x) \).

\[ f'(x) = \frac{d}{dx}[x] \cdot (\ln x - 2) + x \cdot \frac{d}{dx}[\ln x - 2] \]

Calculating the derivatives:

- The derivative of \( x \) is \( 1 \).
- The derivative of \( \ln x \) is \( \frac{1}{x} \) and the derivative of \( -2 \) is \( 0 \).

Thus,

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

### Step 2: Set the derivative to zero to find critical points
To find the critical points, we set \( f'(x) = 0 \):

\[ \ln x - 1 = 0 \implies \ln x = 1 \implies x = e \]

### Step 3: Evaluate the function at the endpoints and critical points
We need to evaluate \( f(x) \) at \( x = 1 \), \( x = e \), and \( x = e^2 \).

1. **At \( x = 1 \)**:
\[ f(1) = 1(\ln 1 - 2) = 1(0 - 2) = -2 \]

2. **At \( x = e \)**:
\[ f(e) = e(\ln e - 2) = e(1 - 2) = e(-1) = -e \]

3. **At \( x = e^2 \)**:
\[ f(e^2) = e^2(\ln e^2 - 2) = e^2(2 - 2) = e^2 \cdot 0 = 0 \]

### Step 4: Determine the greatest and least values
Now we have the values:
- \( f(1) = -2 \)
- \( f(e) = -e \)
- \( f(e^2) = 0 \)

The greatest value is \( 0 \) (at \( x = e^2 \)) and the least value is \( -e \) (at \( x = e \)).

### Step 5: Calculate the difference between the greatest and least values
The difference is:

\[ \text{Difference} = \text{Greatest Value} - \text{Least Value} = 0 - (-e) = 0 + e = e \]

### Final Answer
The difference between the greatest and least values of the function \( f(x) \) on the interval \([1, e^2]\) is \( e \).

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