Given f'(c) = 0, f''(c) = 0, and f'''(c) eq 0, what can be concluded about the point x=c?

We are given that $f'(c) = 0$, $f''(c) = 0$, and $f'''(c) \neq 0$.

• If $f'(c) = 0$, $x=c$ is a critical point.
• If $f''(c) = 0$ as well, the second derivative test is inconclusive for determining local extrema. We must look at higher-order derivatives.
• The higher-order derivative test states that if $f'(c) = 0$, $f''(c) = 0$, ..., $f^{(n-1)}(c) = 0$, but $f^{(n)}(c) \neq 0$:
  • If $n$ is even, then $x=c$ is a local extremum (maximum if $f^{(n)}(c) < 0$, minimum if $f^{(n)}(c) > 0$).
  • If $n$ is odd, then $x=c$ is a point of inflection.

In this case, the first non-zero derivative at $x=c$ is $f'''(c)$, which is the third derivative. The order $n=3$ is an odd number.

Therefore, $x=c$ is a point of inflection.

The correct option is D.

$\boxed{\text{D. A point of inflection}}$