The minimum value of the function $f(x) = \left( \frac{{x^3}}{3} \right) – x$ occurs at $x = 0$.
To find the minimum value, we can take the derivative of $f(x)$ and set it equal to zero. The derivative of $f(x)$ is $f'(x) = x^2 – 1$. Setting $f'(x) = 0$, we get $x^2 = 1$. Solving for $x$, we get $x = \pm 1$.
To see which of these two values of $x$ gives the minimum value of $f(x)$, we can evaluate $f(x)$ at $x = 0$ and $x = \pm 1$. We get $f(0) = 0$ and $f(1) = \frac{1}{3} – 1 = -\frac{2}{3}$. So the minimum value of $f(x)$ occurs at $x = 0$.
We can also see this graphically. The graph of $f(x)$ is a parabola that opens up. The minimum value of a parabola occurs at its vertex. The vertex of the graph of $f(x)$ is at $x = 0$. Therefore, the minimum value of $f(x)$ occurs at $x = 0$.