Find points of local maxima `//`minima of
` f(x) =x^(2) e^(-x)`.

Answer

To find the points of local maxima and minima for the function \( f(x) = x^2 e^{-x} \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \) with respect to \( x \).

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

Using the product rule, we have:

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

Calculating the derivatives:

\[ \frac{d}{dx}(e^{-x}) = -e^{-x} \quad \text{and} \quad \frac{d}{dx}(x^2) = 2x \]

Substituting these into the product rule gives:

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

This simplifies to:

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

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

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

Since \( e^{-x} \) is never zero, we can focus on the equation:

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

Factoring gives:

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

Thus, we have two solutions:

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

### Step 3: Determine the nature of critical points
Next, we need to determine whether these critical points correspond to local maxima, minima, or neither. We can use the second derivative test or evaluate the first derivative around these points.

#### First derivative test:
- For \( x = 0 \):
- Choose a point to the left, say \( x = -1 \):
\[ f'(-1) = e^{1}(2(-1) - (-1)^2) = e( -2 - 1) < 0 \quad (\text{decreasing}) \]
- Choose a point to the right, say \( x = 1 \):
\[ f'(1) = e^{-1}(2(1) - (1)^2) = e^{-1}(2 - 1) > 0 \quad (\text{increasing}) \]
- Therefore, \( x = 0 \) is a local minimum.

- For \( x = 2 \):
- Choose a point to the left, say \( x = 1 \):
\[ f'(1) > 0 \quad (\text{increasing}) \]
- Choose a point to the right, say \( x = 3 \):
\[ f'(3) = e^{-3}(2(3) - (3)^2) = e^{-3}(6 - 9) < 0 \quad (\text{decreasing}) \]
- Therefore, \( x = 2 \) is a local maximum.

### Step 4: Conclusion
The points of local maxima and minima are:
- Local minimum at \( x = 0 \)
- Local maximum at \( x = 2 \)

### Summary of Results
- Local minimum: \( x = 0 \)
- Local maximum: \( x = 2 \)