To find the inflection points of the function $f(x) = x^2 e^{4x}$, we need to determine the second derivative and analyze its sign.

1. First derivative: $f'(x) = 2x e^{4x} + x^2 \cdot 4 e^{4x} = e^{4x} (2x + 4x^2)$

2. Second derivative: Using the product rule: $f''(x) = \frac{d}{dx}(e^{4x}(2x + 4x^2)) = e^{4x}(2 + 8x) + 4e^{4x}(2x + 4x^2)$ This simplifies to: $f''(x) = e^{4x}(2 + 8x + 8x + 16x^2) = e^{4x}(16x^2 + 16x + 2)$ Setting $f''(x) = 0$: $16x^2 + 16x + 2 = 0$ Dividing by 2: $8x^2 + 8x + 1 = 0$ Using the quadratic formula: $x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 8 \cdot 1}}{2 \cdot 8} = \frac{-8 \pm \sqrt{64 - 32}}{16} = \frac{-8 \pm \sqrt{32}}{16} = \frac{-8 \pm 4\sqrt{2}}{16} = \frac{-2 \pm \sqrt{2}}{4}$

   Therefore, the inflection points are: $C = \frac{-2 - \sqrt{2}}{4}, \quad D = \frac{-2 + \sqrt{2}}{4}$

3. Concavity:

   • For $(-\infty, C)$: Choose a test point (e.g., $x = -1$): $f''(-1) = e^{-4}(16(-1)^2 + 16(-1) + 2) = e^{-4}(16 - 16 + 2) = 2 > 0 \quad \text{(Concave Up)}$
   • For $(C, D)$: Choose a test point (e.g., $x = 0$): $f''(0) = e^{0}(16(0)^2 + 16(0) + 2) = 2 > 0 \quad \text{(Concave Up)}$
   • For $(D, \infty)$: Choose a test point (e.g., $x = 1$): $f''(1) = e^{4}(16(1)^2 + 16(1) + 2) = e^{4}(16 + 16 + 2) = 34e^{4} > 0 \quad \text{(Concave Up)}$

Summary:

• Inflection points:

  • $C = \frac{-2 - \sqrt{2}}{4}$
  • $D = \frac{-2 + \sqrt{2}}{4}$

• Concavity:

  • $(-\infty, C)$: Concave Up
  • $(C, D)$: Concave Up
  • $(D, \infty)$: Concave Up

Do you need further details or have any questions? Here are some related questions:

1. How do you interpret inflection points in a graph?
2. What does concavity imply about the behavior of a function?
3. How would the concavity change if the function were modified?
4. Can you provide an example of a function with different concavity?
5. What role does the exponential function play in this context?

Tip: Always verify the sign of the second derivative when determining concavity.