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To find the instantaneous rate of change of the given functions at the specified values, we will compute the derivative and then substitute the given points into the derivative.

### Problem 1: \( y = x^2 - 2x + 1 \) at \( x = 2 \)

1. **Find the derivative**: 
   The derivative of \( y = x^2 - 2x + 1 \) is:
   \[
   \frac{dy}{dx} = 2x - 2
   \]

2. **Substitute \( x = 2 \)**:
   \[
   \frac{dy}{dx} \bigg|_{x=2} = 2(2) - 2 = 4 - 2 = 2
   \]

Thus, the instantaneous rate of change at \( x = 2 \) is **2**. Therefore, the correct answer is \( \boxed{D} \).

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### Problem 2: \( f(x) = x^2 + 2x + 2 \) at \( x = -2 \)

1. **Find the derivative**:
   The derivative of \( f(x) = x^2 + 2x + 2 \) is:
   \[
   \frac{df}{dx} = 2x + 2
   \]

2. **Substitute \( x = -2 \)**:
   \[
   \frac{df}{dx} \bigg|_{x=-2} = 2(-2) + 2 = -4 + 2 = -2
   \]

Thus, the instantaneous rate of change at \( x = -2 \) is **-2**. Therefore, the correct answer is \( \boxed{C} \).

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Do you want further details or explanations? Here are five related questions you could explore:

1. What is the geometric meaning of the instantaneous rate of change?
2. How is the derivative related to the slope of a tangent line?
3. Can you find the second derivative for both functions, and what does it represent?
4. How would you compute the instantaneous rate of change if the function were more complex (e.g., trigonometric or logarithmic)?
5. How does the process change if the function has a higher degree (e.g., cubic or quartic)?

**Tip:** Always check if your function is differentiable before attempting to find the rate of change.

For each problem, find the instantaneous rate of change of the function at the given value: 1) y = x^2 - 2x + 1 at x = 2; 2) f(x) = x^2 + 2x + 2 at x = -2.

Learn how to find the instantaneous rate of change for quadratic functions. This problem involves calculating the derivative and evaluating it at specific points for functions like y = x^2 - 2x + 1 at x = 2 and f(x) = x^2 + 2x + 2 at x = -2. Perfect for high school students studying calculus.

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