The function `f` defined by `f(x)=(x+2)e^(-x)` is

Options

A. decreasing for all `x`
B. decreasing on`(-oo,-1)` and increasing in `(-1,oo)`
C. increasing for all x
D. decreasing in `(-1,oo)` and increasing in `(-oo,-1)`

Explanation

To determine where the function \( f(x) = (x + 2)e^{-x} \) is increasing or decreasing, we need to find its derivative \( f'(x) \) and analyze its sign.

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

Using the product rule, which states that if \( u(x) \) and \( v(x) \) are two functions, then \( (uv)' = u'v + uv' \), we can set:
- \( u = x + 2 \)
- \( v = e^{-x} \)

Now, we calculate the derivatives:
- \( u' = 1 \)
- \( v' = -e^{-x} \) (using the chain rule)

Now applying the product rule:
\[ f'(x) = u'v + uv' = (1)(e^{-x}) + (x + 2)(-e^{-x}) \]
\[ f'(x) = e^{-x} - (x + 2)e^{-x} \]
\[ f'(x) = e^{-x} - (x + 2)e^{-x} = e^{-x}(1 - (x + 2)) \]
\[ f'(x) = e^{-x}(1 - x - 2) = e^{-x}(-x - 1) \]

### Step 2: Analyze the sign of \( f'(x) \)

The derivative can be expressed as:
\[ f'(x) = -e^{-x}(x + 1) \]

Since \( e^{-x} \) is always positive for all \( x \), the sign of \( f'(x) \) depends solely on the term \( -(x + 1) \).

- \( f'(x) > 0 \) when \( -(x + 1) > 0 \) or \( x + 1 < 0 \) which simplifies to \( x < -1 \).
- \( f'(x) < 0 \) when \( -(x + 1) < 0 \) or \( x + 1 > 0 \) which simplifies to \( x > -1 \).

### Step 3: Conclusion

From our analysis:
- The function \( f(x) \) is **increasing** on the interval \( (-\infty, -1) \).
- The function \( f(x) \) is **decreasing** on the interval \( (-1, \infty) \).

Thus, we can conclude:
- The function is increasing for \( x < -1 \) and decreasing for \( x > -1 \).

### Final Answer:
The function \( f(x) = (x + 2)e^{-x} \) is increasing on \( (-\infty, -1) \) and decreasing on \( (-1, \infty) \).

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