The problem given is to find the intervals where the function $f(x) = \frac{1}{3}x^3 - 2x^2 - 5x - 3$ is increasing or decreasing.

Steps to solve:

1. Find the first derivative $f'(x)$: The derivative will help determine the critical points and the behavior of the function. $f'(x) = \frac{d}{dx} \left( \frac{1}{3}x^3 - 2x^2 - 5x - 3 \right)$ Let's differentiate term by term: $f'(x) = x^2 - 4x - 5$

2. Solve for critical points: Set $f'(x) = 0$ to find the critical points: $x^2 - 4x - 5 = 0$ Solve this quadratic equation using the quadratic formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$ $x = \frac{4 \pm \sqrt{16 + 20}}{2}$ $x = \frac{4 \pm \sqrt{36}}{2}$ $x = \frac{4 \pm 6}{2}$ Thus, the solutions are: $x = 5 \quad \text{and} \quad x = -1$

3. Determine the sign of $f'(x)$ in the intervals: The critical points divide the real line into three intervals: $(-\infty, -1)$, $(-1, 5)$, and $(5, \infty)$.

   We will test the sign of $f'(x) = x^2 - 4x - 5$ in each of these intervals:

   • For $x \in (-\infty, -1)$, take $x = -2$: $f'(-2) = (-2)^2 - 4(-2) - 5 = 4 + 8 - 5 = 7 \quad (\text{positive})$

   • For $x \in (-1, 5)$, take $x = 0$: $f'(0) = (0)^2 - 4(0) - 5 = -5 \quad (\text{negative})$

   • For $x \in (5, \infty)$, take $x = 6$: $f'(6) = (6)^2 - 4(6) - 5 = 36 - 24 - 5 = 7 \quad (\text{positive})$

4. Conclusion:

   • The function is increasing on the intervals $(-\infty, -1)$ and $(5, \infty)$.
   • The function is decreasing on the interval $(-1, 5)$.

Would you like more details or explanations? Let me know if you have further questions.

Here are some related questions:

1. How do we classify critical points using the second derivative?
2. What are inflection points, and how are they related to the second derivative?
3. Can we sketch the graph of the cubic function based on the information gathered?
4. How does concavity of the function relate to intervals of increase and decrease?
5. Can we use the discriminant to analyze the behavior of quadratic functions like $f'(x)$?

Tip: Critical points are where the derivative equals zero or does not exist. They help us find maxima, minima, or points of inflection.