We are given the second derivative of a function $f$ as:

$f''(x) = x^2(x - 2)^4(x - 3)^2$

We need to find the x-coordinates of the inflection points, which occur where $f''(x) = 0$ and the concavity changes.

Step 1: Find where $f''(x) = 0$

Set the second derivative equal to zero:

$x^2(x - 2)^4(x - 3)^2 = 0$

For this product to be zero, each factor must be zero:

• $x^2 = 0$ implies $x = 0$
• $(x - 2)^4 = 0$ implies $x = 2$
• $(x - 3)^2 = 0$ implies $x = 3$

Thus, the critical points are $x = 0$, $x = 2$, and $x = 3$.

Step 2: Check for concavity changes

For an inflection point to exist, the concavity must change at these critical points, i.e., $f''(x)$ must change sign.

1. At $x = 0$:

   • $f''(x) = x^2$ is always non-negative and doesn't change sign at $x = 0$. Therefore, $x = 0$ is not an inflection point.

2. At $x = 2$:

   • $(x - 2)^4$ is always non-negative and doesn't change sign at $x = 2$. Therefore, $x = 2$ is not an inflection point.

3. At $x = 3$:

   • $(x - 3)^2$ is also non-negative and doesn't change sign at $x = 3$. Therefore, $x = 3$ is not an inflection point.

Conclusion:

None of the critical points lead to a change in concavity, so there are no inflection points.

Thus, the correct answer is (A) none.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How do you determine if a function has inflection points?
2. What is the relationship between concavity and the second derivative?
3. What is the significance of the sign change in the second derivative for inflection points?
4. How would the process change if the exponents were odd?
5. Can there be inflection points without roots in the second derivative?

Tip: Always remember to check for sign changes when determining inflection points!