Let's solve the problems step by step.

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.