Final answer:
The inflection points of the function f(x) = ax^n, we need to find where the second derivative, f''(x) = an(n-1)x^{n-2}, changes sign. If n > 2, the potential inflection point is at x = 0, assuming it changes the sign of f''(x). If n <= 2, there are no inflection points since the second derivative does not change sign.
Step-by-step explanation:
The question concerns the inflection points of the function f(x) = ax^n. Inflection points occur where the second derivative changes sign, indicating a change in the concavity of the graph.
To determine the inflection points of f(x), we need to find the second derivative f''(x) and solve for x where f''(x) = 0.
Let's calculate the second derivative:
- First derivative: f'(x) = anx^{n-1}.
- Second derivative: f''(x) = an(n-1)x^{n-2}.
If n > 2, we have to set f''(x) = 0 and solve for x to find potential inflection points. However, since x is raised to a power, f''(x) ≠ 0 for any real x, except when x = 0.
Therefore, if x = 0 falls within the domain of f(x) and changes the sign of f''(x), it is an inflection point. If n ≤ 2, the function does not have an inflection point as the second derivative does not change sign.