The problem provides the marginal cost (MC) function $MC = 3Q^2 - 6Q + 4$ and asks for:

1. The total cost (TC) function.
2. The average cost (AC) function.

Here’s how to solve it step by step:

1. Total Cost (TC) Function

The marginal cost $MC$ is the derivative of the total cost $TC$ with respect to $Q$. Thus: $MC = \frac{d(TC)}{dQ}$ To find $TC$, integrate $MC$ with respect to $Q$: $TC = \int MC \, dQ$

Substitute $MC = 3Q^2 - 6Q + 4$: $TC = \int (3Q^2 - 6Q + 4) \, dQ$

Perform the integration: $TC = \int 3Q^2 \, dQ - \int 6Q \, dQ + \int 4 \, dQ$

$TC = Q^3 - 3Q^2 + 4Q + C$

Here, $C$ is the fixed cost or constant of integration.

2. Average Cost (AC) Function

The average cost is the total cost divided by the quantity: $AC = \frac{TC}{Q}$

Substitute $TC = Q^3 - 3Q^2 + 4Q + C$ into $AC$: $AC = \frac{Q^3 - 3Q^2 + 4Q + C}{Q}$

Simplify: $AC = Q^2 - 3Q + 4 + \frac{C}{Q}$

Final Results

1. Total Cost (TC): $TC = Q^3 - 3Q^2 + 4Q + C$

2. Average Cost (AC): $AC = Q^2 - 3Q + 4 + \frac{C}{Q}$

Let me know if you'd like detailed explanations for any part of the solution!

Further Exploration

1. How does the fixed cost $C$ affect the $AC$ function?
2. What is the marginal cost when $Q = 0$?
3. Calculate $AC$ for specific values of $Q$ (e.g., $Q = 1, 2, 3$).
4. What happens to $AC$ as $Q \to \infty$?
5. How does $MC$ change with increasing $Q$?

Tip: Integrals often appear in economics to reverse derivatives; be comfortable applying integration rules to solve cost-related problems.