If f(x) =`e^(x)(x-2)^(2)` then

Options

A. f is increasing in `(-oo,0) ` and `(2,oo) ` an decreasing in (0,2)
B. f is increasing in `(-oo,0) ` and decreasing in `(0,oo)`
C. f is increasing in `(2,00)` and decreasing in `(oo,0)`
D. f is increasing in (0,2) and decreasing in `(-oo,0) ` and `(2,oo)`

Answer

To find the intervals where the function \( f(x) = e^x (x-2)^2 \) is increasing or decreasing, we need to follow these steps:

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

Using the product rule, we differentiate \( f(x) \):

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

Calculating each part:

1. The derivative of \( e^x \) is \( e^x \).
2. The derivative of \( (x-2)^2 \) is \( 2(x-2) \).

Now substituting these into the product rule:

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

Factoring out \( e^x \):

\[ f'(x) = e^x \left( (x-2)^2 + 2(x-2) \right) \]

### Step 2: Simplify the expression

Now we simplify the expression inside the parentheses:

\[ f'(x) = e^x \left( (x-2)^2 + 2(x-2) \right) = e^x \left( (x-2)(x-2 + 2) \right) = e^x \left( (x-2)(x) \right) \]

Thus, we have:

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

### Step 3: Determine the critical points

To find the critical points, we set \( f'(x) = 0 \):

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

Since \( e^x \) is never zero, we focus on the other factors:

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

This gives us the critical points:

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

### Step 4: Test intervals around the critical points

We will test the sign of \( f'(x) \) in the intervals determined by the critical points: \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, 0) \)**:
- Choose \( x = -1 \):
\[ f'(-1) = e^{-1} (-1-2)(-1) = e^{-1} (-3)(-1) = 3e^{-1} > 0 \]
Thus, \( f(x) \) is increasing in \( (-\infty, 0) \).

2. **Interval \( (0, 2) \)**:
- Choose \( x = 1 \):
\[ f'(1) = e^{1} (1-2)(1) = e ( -1)(1) = -e < 0 \]
Thus, \( f(x) \) is decreasing in \( (0, 2) \).

3. **Interval \( (2, \infty) \)**:
- Choose \( x = 3 \):
\[ f'(3) = e^{3} (3-2)(3) = e^{3} (1)(3) = 3e^{3} > 0 \]
Thus, \( f(x) \) is increasing in \( (2, \infty) \).

### Step 5: Conclusion

From the analysis, we conclude:

- \( f(x) \) is increasing on the intervals \( (-\infty, 0] \) and \( [2, \infty) \).
- \( f(x) \) is decreasing on the interval \( (0, 2) \).