To find the inflection points of the function $f(x) = \frac{1}{5}x^5 + 3x^4$, we need to determine where the second derivative $f''(x)$ changes sign. Here are the steps:

1. Find the first derivative $f'(x)$: $f'(x) = x^4 + 12x^3$

2. Find the second derivative $f''(x)$: $f''(x) = 4x^3 + 36x^2$ This can be factored: $f''(x) = 4x^2(x + 9)$

3. Set the second derivative equal to zero to find potential inflection points: $4x^2(x + 9) = 0$ This gives us: $4x^2 = 0 \quad \text{or} \quad x + 9 = 0$ Solving these, we find: $x^2 = 0 \implies x = 0$ $x + 9 = 0 \implies x = -9$

4. Determine the sign change of $f''(x)$:

   • For $x < -9$: Choose $x = -10$: $f''(-10) = 4(-10)^2(-10 + 9) = 4 \times 100 \times (-1) < 0$
   • For $-9 < x < 0$: Choose $x = -1$: $f''(-1) = 4(-1)^2(-1 + 9) = 4 \times 1 \times 8 > 0$
   • For $x > 0$: Choose $x = 1$: $f''(1) = 4(1)^2(1 + 9) = 4 \times 1 \times 10 > 0$

5. Conclusion: The second derivative changes sign at $x = -9$ and $x = 0$, indicating that these points are inflection points.

Inflection Points:

• $x = -9$
• $x = 0$

If you have any questions or would like further details, let me know! Here are some related questions to consider:

1. How do inflection points relate to the concavity of a function?
2. Can you find the coordinates of the inflection points on the graph of $f(x)$?
3. How would you sketch the graph of $f(x)$ using its derivatives?
4. What is the significance of the first derivative test in this context?
5. How does the behavior of $f'(x)$ at the inflection points affect the graph of $f(x)$?

Tip: Always verify sign changes around inflection points by checking the intervals created by the critical points.