To determine the local extrema of the function $f(x) = \frac{8}{3}x^3 - \frac{41}{2}x^2 + 5x + 4$, we need to follow these steps:

1. Calculate the First Derivative: We’ll find $f'(x)$, as setting it to zero will help identify potential critical points, which could correspond to local minima or maxima.

2. Set the Derivative to Zero: Solve $f'(x) = 0$ to find critical points.

3. Calculate the Second Derivative: $f''(x)$ can help determine the concavity at each critical point, allowing us to identify if each critical point is a local maximum or minimum.

Let's go through these steps in detail.

Step 1: First Derivative

Given: $f(x) = \frac{8}{3}x^3 - \frac{41}{2}x^2 + 5x + 4$

Taking the derivative with respect to $x$: $f'(x) = 3 \cdot \frac{8}{3}x^2 - 2 \cdot \frac{41}{2}x + 5 = 8x^2 - 41x + 5$

Step 2: Set $f'(x) = 0$ to Find Critical Points

Now, solve the equation: $8x^2 - 41x + 5 = 0$

Using the quadratic formula: $x = \frac{-(-41) \pm \sqrt{(-41)^2 - 4 \cdot 8 \cdot 5}}{2 \cdot 8}$ $x = \frac{41 \pm \sqrt{1681 - 160}}{16}$ $x = \frac{41 \pm \sqrt{1521}}{16}$ $x = \frac{41 \pm 39}{16}$

This gives: $x = \frac{80}{16} = 5 \quad \text{and} \quad x = \frac{2}{16} = \frac{1}{8}$

So, the critical points are $x = 5$ and $x = \frac{1}{8}$.

Step 3: Second Derivative Test

Now, calculate $f''(x)$ to determine the nature of these critical points.

$f'(x) = 8x^2 - 41x + 5$ Taking the derivative again: $f''(x) = 16x - 41$

• For $x = 5$: $f''(5) = 16 \cdot 5 - 41 = 80 - 41 = 39 \quad (\text{positive, so this is a local minimum})$

• For $x = \frac{1}{8}$: $f''\left(\frac{1}{8}\right) = 16 \cdot \frac{1}{8} - 41 = 2 - 41 = -39 \quad (\text{negative, so this is a local maximum})$

Conclusion

• Local Minimum: $x = 5$
• Local Maximum: $x = \frac{1}{8}$

The answers are:

• Local minimum at $x = 5$.
• Local maximum at $x = \frac{1}{8}$.

Would you like a deeper explanation of any step, or have any questions about this process?

Here are some related questions to further explore:

1. What is the significance of the second derivative test in identifying concavity?
2. How would the procedure differ if we were looking for global extrema instead of local extrema?
3. Could we use a graphical approach to estimate local extrema without calculus?
4. What happens if the second derivative test is inconclusive (i.e., $f''(x) = 0$ at a critical point)?
5. How does the behavior of the function's higher-order derivatives affect the local extrema?

Tip: Always check the endpoints of a function's domain (if bounded) when seeking global extrema, as they might provide the highest or lowest values.