Given the function `f(x)=x^2e^-(2x)`,x>0. Then f(x) has the maximum value equal to a) `e^-1` b) `(2e)^-1 .` c) `e^-2` d) none of these

Explanation

To find the maximum value of the function \( f(x) = x^2 e^{-2x} \) for \( x > 0 \), we will follow these steps:

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

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

Using the product rule, where \( u = x^2 \) and \( v = e^{-2x} \):

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

Calculating \( u' \) and \( v' \):

\[ u' = 2x, \quad v' = -2e^{-2x} \]

Now substituting back into the product rule:

\[ f'(x) = (2x)e^{-2x} + (x^2)(-2e^{-2x}) \]

This simplifies to:

\[ f'(x) = e^{-2x}(2x - 2x^2) \]

Factoring out common terms:

\[ f'(x) = 2e^{-2x}x(1 - x) \]

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

\[ 2e^{-2x}x(1 - x) = 0 \]

This gives us two factors to consider:

1. \( 2e^{-2x} = 0 \) (not possible since \( e^{-2x} > 0 \) for all \( x \))
2. \( x(1 - x) = 0 \)

From \( x(1 - x) = 0 \), we get:

\[ x = 0 \quad \text{or} \quad x = 1 \]

Since we are considering \( x > 0 \), we have \( x = 1 \).

### Step 3: Determine if it is a maximum
To confirm that \( x = 1 \) is a maximum, we can use the second derivative test or analyze the sign of \( f'(x) \) around \( x = 1 \).

Calculating the second derivative \( f''(x) \) is complex, so we can check the sign of \( f'(x) \):

- For \( x < 1 \), \( f'(x) > 0 \) (function is increasing).
- For \( x > 1 \), \( f'(x) < 0 \) (function is decreasing).

This indicates that \( x = 1 \) is indeed a maximum.

### Step 4: Calculate the maximum value
Now we find the maximum value of \( f(x) \) at \( x = 1 \):

\[ f(1) = 1^2 e^{-2 \cdot 1} = e^{-2} \]

### Conclusion
Thus, the maximum value of \( f(x) \) is:

\[ \boxed{e^{-2}} \]