To solve this problem, let’s follow the steps needed to analyze the function $f(x) = (3x + 1)e^x$:

1. Find the critical numbers of $f$ and $f'$:

   • A critical point occurs where $f'(x) = 0$ or $f'(x)$ is undefined.

2. Calculate $f'(x)$:

   • Use the product rule: if $f(x) = u(x)v(x)$ with $u(x) = 3x + 1$ and $v(x) = e^x$, then $f'(x) = u'(x)v(x) + u(x)v'(x).$
   • Compute $u'(x) = 3$ and $v'(x) = e^x$, so: $f'(x) = 3e^x + (3x + 1)e^x = e^x(3x + 4).$

3. Solve $f'(x) = 0$:

   • Set $e^x(3x + 4) = 0$.
   • Since $e^x \neq 0$, we only need to solve $3x + 4 = 0$: $x = -\frac{4}{3}.$
   • This is one of the critical points.

4. Determine Concavity and Inflection Points:

   • To analyze concavity, calculate $f''(x)$.
   • Differentiate $f'(x) = e^x(3x + 4)$ again using the product rule: $f''(x) = e^x(3x + 4) + e^x \cdot 3 = e^x(3x + 7).$
   • Set $f''(x) = 0$ to find potential inflection points: $e^x(3x + 7) = 0.$
   • Solving $3x + 7 = 0$: $x = -\frac{7}{3}.$
   • So, $x = -\frac{7}{3}$ is a potential inflection point.

5. Behavior Analysis:

   • With the critical points $a = -\frac{4}{3}$ and $b = -\frac{7}{3}$, analyze each interval:
     • $x < a$: Decreasing and concave downward.
     • $a < x < b$: Increasing and concave downward.
     • $x > b$: Increasing and concave upward.

6. Limit Behavior at $\pm \infty$:

   • As $x \to -\infty$, $f(x) \to 0$.
   • As $x \to \infty$, $f(x) \to \infty$.

These steps should allow you to fill in the table:

• $a = -\frac{4}{3}$, $b = -\frac{7}{3}$
• Behavior at each interval:
  • $x < a$: Decreasing, concave downward.
  • $a < x < b$: Increasing, concave downward.
  • $x > b$: Increasing, concave upward.
• At $-\infty$ and $+\infty$: "Other"