The average cost function $A(x)$ is given by:

$A(x) = \frac{C(x)}{x} = \frac{x^3 - 14x^2 + 163x + 128}{x}$

Simplifying the expression for $A(x)$, we get:

$A(x) = x^2 - 14x + 163 + \frac{128}{x}$

To determine when $A(x)$ is increasing or decreasing, we need to find its derivative $A'(x)$ and analyze the sign of this derivative.

1. First, differentiate $A(x)$:

$A'(x) = \frac{d}{dx} \left( x^2 - 14x + 163 + \frac{128}{x} \right)$

Using basic differentiation rules:

$A'(x) = 2x - 14 - \frac{128}{x^2}$

2. Set $A'(x) = 0$ to find the critical points:

$2x - 14 - \frac{128}{x^2} = 0$

Multiplying the equation by $x^2$ to eliminate the fraction:

$2x^3 - 14x^2 - 128 = 0$

Divide the equation by 2:

$x^3 - 7x^2 - 64 = 0$

Now, solve this cubic equation:

$(x - 8)(x^2 + x + 8) = 0$

This gives us $x = 8$ from the factor $(x - 8)$, and the quadratic part $x^2 + x + 8 = 0$ has no real solutions because its discriminant is negative.

So, the critical point is $x = 8$.

3. Determine where $A'(x)$ is positive (increasing) or negative (decreasing) by testing values of $x$ around $x = 8$.

• For $x > 8$, choose $x = 9$:

$A'(9) = 2(9) - 14 - \frac{128}{9^2} = 18 - 14 - \frac{128}{81} \approx 4 - 1.58 = 2.42$

Since $A'(9) > 0$, $A(x)$ is increasing for $x > 8$.

• For $0 < x < 8$, choose $x = 7$:

$A'(7) = 2(7) - 14 - \frac{128}{7^2} = 14 - 14 - \frac{128}{49} \approx 0 - 2.61 = -2.61$

Since $A'(7) < 0$, $A(x)$ is decreasing for $0 < x < 8$.

4. Conclusion:

• $A(x)$ is decreasing for $0 < x < 8$.
• A(x) is **increasing** for \( x > 8.

Thus, the correct answer is:

Increasing: $x > 8$; Decreasing: $0 < x < 8$

Would you like more details on this?

Here are 5 related questions:

1. How do you find the critical points of a function?
2. How do you determine intervals where a function is increasing or decreasing?
3. What are the properties of cubic functions like the one in this problem?
4. How do you interpret the behavior of cost functions in economics?
5. How can you solve cubic equations more easily?

Tip: When dealing with average cost functions, look for places where the derivative changes sign to find the intervals of increasing and decreasing behavior.