Answer
The minimum value is $-14$ attained when $x=-\sqrt[3] 2$.
Work Step by Step
As suggested, we use the substitution $t=x^3$ to get $f(x) = 2 + 16t+4t^2$ which is a quadratic function of the form $f(t) = at^2+bt+c$.
To find the minimum value, we find the vertex using the methods outlined in the chapter; mainly, $h=-\frac{b}{2a}$ and $k=f(h)$ with the vertex being the point $(h, k).$
We plug the values to get $h=-2$ and $k=-14.$ We note that this would represent the set of values of $f(t)$ rather than the required $f(x)$. To get the values for $f(x)$, we plug in $t=x^3$ to get that $x=\sqrt[3]{t}$ and hence the minimum value of $f(x)$ is $-14$ attained when $x=-\sqrt[3]{2}$.
Attached below is a graph illustrating the curves mentioned above and the respective minimum points.
