Inflection points occur where $\displaystyle f\prime \prime (x) = 0$
If $\displaystyle f\prime \prime (x) = 0$ at $\displaystyle x = -1$ and $\displaystyle x = 3$, for example,
then your intervals are $\displaystyle ( - \infty , - 1), ( - 1,3), (3, + \infty )$
You can determine the concavity of an interval by checking any value within that interval using $\displaystyle f\prime \prime (x)$
For the interval $\displaystyle ( - \infty , - 1)$, you could use the number -10
If $\displaystyle f\prime \prime (-10)$ is positive, then that interval is concave up.
If $\displaystyle f\prime \prime (-10)$ is negative, then that interval is concave down.
Also, if $\displaystyle f\prime \prime (x)$ is negative on both sides or positive on both sides, then there is no inflection point at x.
$\displaystyle \begin{array}{l}
f(x) = {x^4} - 4{x^3} \\
f\prime (x) = 4{x^3} - 12{x^2} \\
f\prime \prime (x) = 12{x^2} - 24x = 12x(x - 2) \\ \end{array}$
$\displaystyle f\prime \prime (x) = 0$ @ $\displaystyle x = 0, 2 \\ $
Can you figure it out from there?