To analyze the function $f(x) = x^3 - 18x^2 + 96x - 80$, let's follow the steps:

1. Find the critical values:

To find the critical values, first compute the derivative: $f'(x) = 3x^2 - 36x + 96$ Setting the derivative equal to zero: $3x^2 - 36x + 96 = 0$ Dividing by 3: $x^2 - 12x + 32 = 0$ Using the quadratic formula: $x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm 4}{2}$ Thus: $x = 8 \quad \text{and} \quad x = 4$ Critical values: $x = 4$ and $x = 8$

2. Test for concavity:

Next, find the second derivative: $f''(x) = 6x - 36$ Evaluating at the critical points:

• For $x = 4$: $f''(4) = 6(4) - 36 = 24 - 36 = -12 \quad (\text{concave down, local max})$
• For $x = 8$: $f''(8) = 6(8) - 36 = 48 - 36 = 12 \quad (\text{concave up, local min})$

3. Evaluate the function at the critical values:

• For $x = 4$: $f(4) = 4^3 - 18(4^2) + 96(4) - 80 = 64 - 288 + 384 - 80 = 80$
• For $x = 8$: $f(8) = 8^3 - 18(8^2) + 96(8) - 80 = 512 - 1152 + 768 - 80 = 48$

Summary:

• Critical values: $x = 4$ (max, $f(4) = 80$), $x = 8$ (min, $f(8) = 48$).

Would you like more details on any part? Here are some follow-up questions:

1. How do we interpret the results of the concavity test?
2. What do the relative maxima and minima indicate about the function's behavior?
3. How would the graph of the function look around these critical values?
4. Can we find inflection points for this function?
5. What would happen if we changed the coefficients in the function?

Tip: Always check the behavior of the function beyond critical points for a complete understanding.