given f(x)= (x-3)^2 e^-2x, find the critical values of f, the intervals where f is increasing/decreasing, and the x values of any local extrema of f (specify the type).

To solve the problem, we will follow these steps:

1. **Find the critical values**: Critical values occur where \( f'(x) = 0 \) or is undefined.
2. **Determine where \( f(x) \) is increasing or decreasing**: This depends on the sign of \( f'(x) \) on different intervals.
3. **Find the x-values of any local extrema**: Use the first derivative test to classify the critical points as either local minima or maxima.

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

Given the function \( f(x) = (x-3)^2 e^{-2x} \), we will apply the product rule and the chain rule. Let:
- \( u(x) = (x-3)^2 \)
- \( v(x) = e^{-2x} \)

The product rule states:
\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]

#### Derivatives of \( u(x) \) and \( v(x) \):
- \( u'(x) = 2(x-3) \)
- \( v'(x) = \frac{d}{dx}[e^{-2x}] = -2e^{-2x} \)

Now, applying the product rule:
\[
f'(x) = 2(x-3)e^{-2x} + (x-3)^2(-2e^{-2x})
\]
Factor out \( e^{-2x} \):
\[
f'(x) = e^{-2x} \left[ 2(x-3) - 2(x-3)^2 \right]
\]
Simplify inside the brackets:
\[
f'(x) = e^{-2x} \left[ 2(x-3) \left(1 - (x-3)\right) \right]
\]
\[
f'(x) = e^{-2x} \left[ 2(x-3)(4 - x) \right]
\]

### Step 2: Find the critical points

To find the critical points, set \( f'(x) = 0 \). Since \( e^{-2x} \neq 0 \) for all \( x \), we solve:
\[
2(x-3)(4 - x) = 0
\]

This gives two solutions:
\[
x - 3 = 0 \quad \text{or} \quad 4 - x = 0
\]
So, \( x = 3 \) and \( x = 4 \).

### Step 3: Determine where \( f(x) \) is increasing or decreasing

To determine the behavior of \( f'(x) \), analyze its sign on the intervals determined by the critical points: \( (-\infty, 3) \), \( (3, 4) \), and \( (4, \infty) \).

1. **On \( (-\infty, 3) \):**
   - Pick a test point, say \( x = 2 \):
     \[
     f'(2) = e^{-4} \cdot 2(2-3)(4-2) = e^{-4} \cdot 2(-1)(2) = -4e^{-4} < 0
     \]
   - So, \( f(x) \) is decreasing on \( (-\infty, 3) \).

2. **On \( (3, 4) \):**
   - Pick a test point, say \( x = 3.5 \):
     \[
     f'(3.5) = e^{-7} \cdot 2(3.5-3)(4-3.5) = e^{-7} \cdot 2(0.5)(0.5) = 0.5e^{-7} > 0
     \]
   - So, \( f(x) \) is increasing on \( (3, 4) \).

3. **On \( (4, \infty) \):**
   - Pick a test point, say \( x = 5 \):
     \[
     f'(5) = e^{-10} \cdot 2(5-3)(4-5) = e^{-10} \cdot 2(2)(-1) = -4e^{-10} < 0
     \]
   - So, \( f(x) \) is decreasing on \( (4, \infty) \).

### Step 4: Find the x-values of any local extrema

Using the first derivative test:
- At \( x = 3 \), \( f(x) \) changes from decreasing to increasing. Thus, \( x = 3 \) is a **local minimum**.
- At \( x = 4 \), \( f(x) \) changes from increasing to decreasing. Thus, \( x = 4 \) is a **local maximum**.

### Summary of Results:
- **Critical points**: \( x = 3 \) and \( x = 4 \)
- **Intervals of increase/decrease**:
  - Decreasing on \( (-\infty, 3) \) and \( (4, \infty) \)
  - Increasing on \( (3, 4) \)
- **Local extrema**:
  - Local minimum at \( x = 3 \)
  - Local maximum at \( x = 4 \)

---

Would you like further details or clarifications?

Here are 5 related questions:
1. How do second derivatives help in identifying concavity and points of inflection?
2. How does the product rule differ from the quotient rule in derivative calculations?
3. How would you find the global extrema of \( f(x) \) on a closed interval?
4. What happens if you encounter a critical point where the first derivative doesn't change signs?
5. How do you find the asymptotes of functions involving exponential terms?

**Tip**: To confirm the type of extrema, you can also use the second derivative test!

Given f(x)= (x-3)^2 e^-2x, find the critical values of f, the intervals where f is increasing/decreasing, and the x values of any local extrema of f (specify the type).

Learn how to find the critical values, intervals of increase/decrease, and local extrema for the function f(x) = (x-3)^2 e^-2x using the first derivative test. This detailed solution applies calculus techniques including the product and chain rules, and is ideal for students in Grades 11-12 studying advanced calculus.

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