On [1,e] the greatest value of `x^2log_ex`, is

Options

A. `e^2`
B. `1/2log(1/sqrte)`
C. `e^2logsqrte`
D. e

Answer

To find the greatest value of the function \( f(x) = x^2 \log_e x \) on the interval \([1, e]\), we will follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^2 \log_e x) \]

Using the product rule, we have:

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

Thus, we can simplify this to:

\[ f'(x) = x(2 \log_e x + 1) \]

### Step 2: Find critical points
Next, we set the derivative equal to zero to find critical points.

\[ f'(x) = 0 \implies x(2 \log_e x + 1) = 0 \]

Since \( x = 0 \) is not in our interval \([1, e]\), we focus on the term:

\[ 2 \log_e x + 1 = 0 \]

Solving for \( x \):

\[ 2 \log_e x = -1 \implies \log_e x = -\frac{1}{2} \implies x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

### Step 3: Check if the critical point is in the interval
The value \( \frac{1}{\sqrt{e}} \) is approximately \( 0.606 \), which is less than 1. Therefore, it is not in the interval \([1, e]\).

### Step 4: Evaluate the function at the endpoints
Since there are no critical points in the interval, we evaluate the function at the endpoints of the interval:

1. At \( x = 1 \):
\[ f(1) = 1^2 \log_e 1 = 1 \cdot 0 = 0 \]

2. At \( x = e \):
\[ f(e) = e^2 \log_e e = e^2 \cdot 1 = e^2 \]

### Step 5: Determine the greatest value
Now we compare the values obtained:

- \( f(1) = 0 \)
- \( f(e) = e^2 \)

Since \( e^2 > 0 \), the greatest value of \( f(x) \) on the interval \([1, e]\) is:

\[ \text{Greatest value} = e^2 \]

### Conclusion
Thus, the greatest value of \( x^2 \log_e x \) on the interval \([1, e]\) is \( e^2 \).

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