The range of the function `f(x)=x^(2)ln(x)" for "x in [1, e]" is " [a, b]`, where `a+b` is equal to

Options

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

Answer

To find the range of the function \( f(x) = x^2 \ln(x) \) for \( x \) in the interval \([1, e]\), we will follow these steps:

### Step 1: Determine the function and its domain
The function given is:
\[ f(x) = x^2 \ln(x) \]
The domain specified is \( x \in [1, e] \).

### Step 2: Find the derivative of the function
To determine if the function is increasing or decreasing, we need to find its derivative:
\[ f'(x) = \frac{d}{dx}(x^2 \ln(x)) \]
Using the product rule:
\[ f'(x) = x^2 \cdot \frac{1}{x} + \ln(x) \cdot 2x = x + 2x \ln(x) \]

### Step 3: Analyze the derivative
We need to check if \( f'(x) \) is positive in the interval \([1, e]\):
\[ f'(x) = x(1 + 2 \ln(x)) \]
Since \( x > 0 \) in the interval, we need to analyze \( 1 + 2 \ln(x) \):
- At \( x = 1 \), \( \ln(1) = 0 \) so \( f'(1) = 1 + 0 = 1 > 0 \).
- At \( x = e \), \( \ln(e) = 1 \) so \( f'(e) = 1 + 2 \cdot 1 = 3 > 0 \).

Since \( f'(x) > 0 \) for all \( x \in [1, e] \), the function \( f(x) \) is increasing in this interval.

### Step 4: Calculate the function values at the endpoints
Now we will find the values of the function at the endpoints of the interval:
- At \( x = 1 \):
\[ f(1) = 1^2 \ln(1) = 1 \cdot 0 = 0 \]
- At \( x = e \):
\[ f(e) = e^2 \ln(e) = e^2 \cdot 1 = e^2 \]

### Step 5: Determine the range
Since the function is increasing, the range of \( f(x) \) as \( x \) goes from \( 1 \) to \( e \) is:
\[ f(x) \in [f(1), f(e)] = [0, e^2] \]

### Step 6: Find \( a + b \)
Here, \( a = 0 \) and \( b = e^2 \). Therefore:
\[ a + b = 0 + e^2 = e^2 \]

### Final Answer
Thus, the value of \( a + b \) is:
\[ \boxed{e^2} \]