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

Answer

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

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

Using the product rule, where \( u = x^2 \) and \( v = e^{-x} \):
\[ f'(x) = u'v + uv' \]
Calculating \( u' \) and \( v' \):
- \( u' = 2x \)
- \( v' = -e^{-x} \)

Now applying the product rule:
\[ f'(x) = (2x)e^{-x} + (x^2)(-e^{-x}) = e^{-x}(2x - x^2) \]

### Step 2: Set the first derivative to zero
To find the critical points, we need to set the first derivative equal to zero:
\[ f'(x) = e^{-x}(2x - x^2) = 0 \]

Since \( e^{-x} \) is never zero for any real \( x \), we can simplify this to:
\[ 2x - x^2 = 0 \]
Factoring out \( x \):
\[ x(2 - x) = 0 \]

### Step 3: Solve for critical points
Setting each factor to zero gives us:
1. \( x = 0 \)
2. \( 2 - x = 0 \) which gives \( x = 2 \)

Thus, the critical points are \( x = 0 \) and \( x = 2 \).

### Step 4: Determine the nature of the critical points
To classify the critical points, we can use the first derivative test. We will evaluate \( f'(x) \) in the intervals around the critical points.

- Choose a test point in the interval \( (-\infty, 0) \), say \( x = -1 \):
\[ f'(-1) = e^{1}(2(-1) - (-1)^2) = e( -2 - 1) < 0 \quad (\text{negative}) \]

- Choose a test point in the interval \( (0, 2) \), say \( x = 1 \):
\[ f'(1) = e^{-1}(2(1) - (1)^2) = e^{-1}(2 - 1) > 0 \quad (\text{positive}) \]

- Choose a test point in the interval \( (2, \infty) \), say \( x = 3 \):
\[ f'(3) = e^{-3}(2(3) - (3)^2) = e^{-3}(6 - 9) < 0 \quad (\text{negative}) \]

### Step 5: Conclusion
From the sign changes of \( f'(x) \):
- At \( x = 0 \), \( f'(x) \) changes from negative to positive, indicating a local minimum.
- At \( x = 2 \), \( f'(x) \) changes from positive to negative, indicating a local maximum.

Thus, we conclude:
- Local minimum at \( x = 0 \)
- Local maximum at \( x = 2 \)