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To find the critical numbers for the function \( f(x) = \frac{4}{3}x^3 - 4x^2 - 60x - 20 \), we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \).
The derivative, \( f'(x) \), gives us the slope of the tangent line at any point on the curve. We need to find \( f'(x) \) first:
\[
f'(x) = \frac{d}{dx} \left( \frac{4}{3}x^3 - 4x^2 - 60x - 20 \right) = 4x^2 - 8x - 60.
\]

### Step 2: Set the derivative equal to zero to find critical points.
We solve:
\[
4x^2 - 8x - 60 = 0.
\]

### Step 3: Solve this quadratic equation.
We can simplify by dividing all terms by 4:
\[
x^2 - 2x - 15 = 0.
\]
Factor the quadratic equation:
\[
(x - 5)(x + 3) = 0.
\]
This gives the solutions:
\[
x = 5 \quad \text{and} \quad x = -3.
\]

Thus, the critical numbers are \( x = 5 \) and \( x = -3 \).

Would you like details on the next parts (b) and (c), or have any questions about this solution?

### Further Questions:
1. How do you determine if a critical point is a maximum, minimum, or neither?
2. How can the second derivative be used to confirm concavity at a critical point?
3. What are inflection points, and how do they relate to the second derivative?
4. Can critical points occur at endpoints of a function's domain?
5. How would this process change if the function were defined piecewise?

**Tip:** When solving a quadratic equation, factoring is often simpler if the coefficients are small. If not, using the quadratic formula is more reliable.

For the function below, find (a) the critical numbers; (b) the open intervals where the function is increasing; and (c) the open intervals where it is decreasing.

f(x) = (4/3)x^3 - 4x^2 - 60x - 20

Determine the critical numbers.

Discover how to find the critical numbers of the function f(x) = (4/3)x^3 - 4x^2 - 60x - 20. This solution uses calculus techniques, such as taking derivatives and solving quadratic equations, to determine the points where the function's slope is zero or undefined.

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