The question asks for the following:

1. Identify intervals where $f(x)$ is increasing and decreasing.
2. Determine intervals where $f(x)$ is concave up and concave down.
3. Graph $f(x)$ to verify your results.

The function is:

$f(x) = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x$

Step-by-Step Solution

Step 1: First Derivative ($f'(x)$)

To find intervals of increase and decrease, compute $f'(x)$: $f'(x) = x^2 - x - 6$

Factorize: $f'(x) = (x - 3)(x + 2)$

Critical points occur where $f'(x) = 0$: $x = 3 \quad \text{and} \quad x = -2$

Test Intervals for $f'(x)$

Split the number line into intervals based on critical points: $(-\infty, -2)$, $(-2, 3)$, $(3, \infty)$.

• For $x \in (-\infty, -2)$: Pick $x = -3$: $f'(-3) = (-3 - 3)(-3 + 2) = (-6)(-1) = 6 \quad (\text{positive, increasing})$

• For $x \in (-2, 3)$: Pick $x = 0$: $f'(0) = (0 - 3)(0 + 2) = (-3)(2) = -6 \quad (\text{negative, decreasing})$

• For $x \in (3, \infty)$: Pick $x = 4$: $f'(4) = (4 - 3)(4 + 2) = (1)(6) = 6 \quad (\text{positive, increasing})$

Conclusion for $f'(x)$

• $f(x)$ is increasing on $(-\infty, -2) \cup (3, \infty)$.
• $f(x)$ is decreasing on $(-2, 3)$.

Step 2: Second Derivative ($f''(x)$)

To find concavity, compute $f''(x)$: $f''(x) = 2x - 1$

Set $f''(x) = 0$ to find inflection points: $2x - 1 = 0 \implies x = \frac{1}{2}$

Test Intervals for $f''(x)$

Split the number line into intervals: $(-\infty, \frac{1}{2})$, $(\frac{1}{2}, \infty)$.

• For $x \in (-\infty, \frac{1}{2})$: Pick $x = 0$: $f''(0) = 2(0) - 1 = -1 \quad (\text{negative, concave down})$

• For $x \in (\frac{1}{2}, \infty)$: Pick $x = 1$: $f''(1) = 2(1) - 1 = 1 \quad (\text{positive, concave up})$

Conclusion for $f''(x)$

• $f(x)$ is concave down on $(-\infty, \frac{1}{2})$.
• $f(x)$ is concave up on $(\frac{1}{2}, \infty)$.

Step 3: Graph $f(x)$

Using the above results:

• Critical points: $x = -2, 3$.
• Inflection point: $x = \frac{1}{2}$.
• Behavior:
  • Increasing: $(-\infty, -2)$, $(3, \infty)$.
  • Decreasing: $(-2, 3)$.
  • Concave up: $(\frac{1}{2}, \infty)$.
  • Concave down: $(-\infty, \frac{1}{2})$.

Would you like me to draw the graph for you or provide any additional clarifications?

Questions for Deeper Understanding

1. How does the sign of $f'(x)$ determine increasing or decreasing intervals?
2. What does the second derivative tell us about the shape of the graph?
3. How do critical points and inflection points influence the graph's shape?
4. Can you determine turning points from the graph of $f(x)$?
5. Why is concavity important for understanding optimization problems?

Tip: Always verify results by substituting key points into the function to confirm behavior.