Answer
$f$ does not have a local maximum or a local minimum at $c$. However, $f$ has an inflection point at $c$ since the concavity changes at $c$.
Work Step by Step
$f'''$ is continuous and $f'''(c) \gt 0$
Thus, there is a neighborhood of $c$ such that if $~~c-\delta \lt x \lt c+\delta~~$, then $f'''(x) \gt 0$
Then $f''$ is increasing in this neighborhood of $c$
If $c-\delta \lt x \lt c$, $f''(x) \lt 0$
$f$ is concave down on the interval $(c-\delta, c)$
If $c \lt x \lt c+\delta$, $f''(x) \gt 0$
$f$ is concave up on the interval $(c, c+\delta)$
Therefore, $f$ does not have a local maximum or a local minimum at $c$. However, $f$ has an inflection point at $c$ since the concavity changes at $c$
An example of this situation is $f(x) = x^3$ where $c = 0$
$f'(0) = 3(0)^2 = 0$
$f''(0) = 6(0) = 0$
$f'''(0) = 6 \gt 0$
Note that $f(x) = x^3$ does not have a local maximum or a local minimum at $x=0$, but $f(x) = x^3$ has an inflection point at $x=0$
