Answer
$(0,0)$ is not an inflection point of the graph $f$.
Work Step by Step
$f(x) = x^4$
$f'(x) = 4x^3$
$f''(x) = 12x^2$
$f''(x) = 0$ when $x = 0$
$f(0) = 0$, so the point $(0,0)$ is a point on the graph of $f(x)$
However:
$f''(x) \gt 0$ when $x \lt 0$
$f''(x) \gt 0$ when $x \gt 0$
Therefore, $f(x)$ is concave upward. At the point $(0,0)$, the graph does not change from concave upward to concave downward (or vice versa). Therefore, $(0,0)$ is not an inflection point of the graph $f$.
