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Let's solve the problem step by step. We are given the cost function \( C(x) = 7000 + 5x + 0.5x^2 \), where \( C(x) \) represents the cost of producing \( x \) units of a commodity.

### Part (a) - Average Rate of Change

The average rate of change of \( C \) with respect to \( x \) between two points \( x_1 \) and \( x_2 \) is given by the formula:

\[
\text{Average rate of change} = \frac{C(x_2) - C(x_1)}{x_2 - x_1}
\]

#### (i) From \( x = 100 \) to \( x = 105 \)

Let's calculate the values of \( C(100) \) and \( C(105) \).

1. \( C(100) = 7000 + 5(100) + 0.5(100^2) = 7000 + 500 + 0.5(10000) = 7000 + 500 + 5000 = 12500 \)
2. \( C(105) = 7000 + 5(105) + 0.5(105^2) = 7000 + 525 + 0.5(11025) = 7000 + 525 + 5512.5 = 13037.5 \)

Now, apply the average rate of change formula:

\[
\text{Average rate of change} = \frac{13037.5 - 12500}{105 - 100} = \frac{537.5}{5} = 107.5
\]

So, the average rate of change from \( x = 100 \) to \( x = 105 \) is **107.5**.

#### (ii) From \( x = 100 \) to \( x = 101 \)

Similarly, let's calculate the values of \( C(101) \).

1. \( C(101) = 7000 + 5(101) + 0.5(101^2) = 7000 + 505 + 0.5(10201) = 7000 + 505 + 5100.5 = 12605.5 \)

Now, apply the average rate of change formula:

\[
\text{Average rate of change} = \frac{12605.5 - 12500}{101 - 100} = \frac{105.5}{1} = 105.5
\]

So, the average rate of change from \( x = 100 \) to \( x = 101 \) is **105.5**.

### Part (b) - Instantaneous Rate of Change (Marginal Cost)

The instantaneous rate of change of \( C \) at \( x = 100 \) is found by taking the derivative of \( C(x) \) and evaluating it at \( x = 100 \). The derivative of \( C(x) \) is:

\[
C'(x) = \frac{d}{dx} \left(7000 + 5x + 0.5x^2 \right) = 0 + 5 + x = 5 + x
\]

Now, evaluate \( C'(100) \):

\[
C'(100) = 5 + 100 = 105
\]

So, the instantaneous rate of change (marginal cost) at \( x = 100 \) is **105**.

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Would you like any additional details or clarification on any of the steps? Here are five related questions to expand on this problem:

1. How does the marginal cost function change for different values of \( x \)?
2. What is the economic interpretation of the average rate of change in this context?
3. How would the cost function change if the coefficient of \( x^2 \) were different?
4. What happens to the marginal cost as \( x \) increases?
5. How would you interpret the marginal cost if it were negative?

**Tip:** The marginal cost function \( C'(x) \) gives a direct estimate of the cost of producing one more unit of the commodity at a given production level \( x \).

Find the average and instantaneous rate of change of the cost function C(x) = 7000 + 5x + 0.5x^2 for different intervals of x.

This problem involves calculating the average and instantaneous rates of change for the cost function C(x) = 7000 + 5x + 0.5x^2. It covers key calculus concepts, including the difference quotient and derivatives, suitable for high school students.

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