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### Given function:
\[ P(x) = 5x^2 - 6x + 6 \]

### a. Average rate of change from \( x = 3 \) to \( x = 5 \):
\[
\text{Average rate} = \frac{P(5) - P(3)}{5 - 3}
\]

Calculate \( P(5) \) and \( P(3) \):

\[ P(5) = 5(5)^2 - 6(5) + 6 = 125 - 30 + 6 = 101 \]
\[ P(3) = 5(3)^2 - 6(3) + 6 = 45 - 18 + 6 = 33 \]

\[ \text{Average rate} = \frac{101 - 33}{2} = \frac{68}{2} = 34 \]

**Answer:** 34 per item

### b. Average rate of change from \( x = 3 \) to \( x = 4 \):
\[
\text{Average rate} = \frac{P(4) - P(3)}{4 - 3}
\]

Calculate \( P(4) \):

\[ P(4) = 5(4)^2 - 6(4) + 6 = 80 - 24 + 6 = 62 \]

\[ \text{Average rate} = \frac{62 - 33}{1} = 29 \]

**Answer:** 29 per item

### c. Instantaneous rate of change at \( x = 3 \) (marginal profit):
Find the derivative \( P'(x) \):

\[ P'(x) = \frac{d}{dx}(5x^2 - 6x + 6) = 10x - 6 \]

Evaluate at \( x = 3 \):

\[ P'(3) = 10(3) - 6 = 30 - 6 = 24 \]

**Answer:** 24 per item

### Interpretation:
At \( x = 3 \), the profit is increasing at the rate of 24 per item.

**Correct choice:** B

### d. Marginal profit at \( x = 5 \):
\[ P'(5) = 10(5) - 6 = 50 - 6 = 44 \]

**Answer:** 44 per item

If you have any questions or need further explanation, feel free to ask!

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1. How do you find the derivative of a polynomial function?
2. What is the difference between average rate and instantaneous rate?
3. How do you interpret marginal profit?
4. What is the significance of a negative rate of change?
5. How do you calculate profit from a quadratic function?

**Tip:** Remember that the derivative represents the slope of the tangent line to the curve at any point, indicating the instantaneous rate of change.

Learn how to calculate average and instantaneous rates of change in polynomial functions using specific examples. This problem involves applying the derivative of a polynomial function and interpreting the results for profit analysis. Suitable for high school students.

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