The function `f(x)=x^(2)e^(x)` has maximum value

Options

A. 0
B. e
C. `(1)/(e )`
D. `(4)/(e^(2))`

Answer

To find the maximum value of the function \( f(x) = x^2 e^x \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)

Using the product rule for differentiation, where \( u = x^2 \) and \( v = e^x \):

\[ f'(x) = u'v + uv' = (2x)e^x + (x^2)(e^x) = e^x(2x + x^2) \]

### Step 2: Set the first derivative to zero to find critical points

We set \( f'(x) = 0 \):

\[ e^x(2x + x^2) = 0 \]

Since \( e^x \) is never zero, we can simplify to:

\[ 2x + x^2 = 0 \]

Factoring out \( x \):

\[ x(2 + x) = 0 \]

This gives us two solutions:

\[ x = 0 \quad \text{or} \quad x = -2 \]

### Step 3: Determine the nature of the critical points using the second derivative test

Now we find the second derivative \( f''(x) \):

Using the product rule again on \( f'(x) = e^x(2x + x^2) \):

\[ f''(x) = e^x(2x + x^2)' + (e^x)'(2x + x^2) \]
\[ = e^x(2 + 2x) + e^x(2x + x^2) = e^x(2 + 2x + 2x + x^2) = e^x(x^2 + 4x + 2) \]

### Step 4: Evaluate the second derivative at the critical points

1. For \( x = 0 \):
\[ f''(0) = e^0(0^2 + 4(0) + 2) = 2 > 0 \quad \text{(local minimum)} \]

2. For \( x = -2 \):
\[ f''(-2) = e^{-2}((-2)^2 + 4(-2) + 2) = e^{-2}(4 - 8 + 2) = e^{-2}(-2) < 0 \quad \text{(local maximum)} \]

### Step 5: Find the maximum value

Now we calculate \( f(-2) \):

\[ f(-2) = (-2)^2 e^{-2} = 4 e^{-2} \]

### Conclusion

The maximum value of the function \( f(x) = x^2 e^x \) occurs at \( x = -2 \) and is given by:

\[ \text{Maximum value} = \frac{4}{e^2} \]