The mean value theorem states that if a function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) – f(a)}{b – a}$.
In this case, we have $f(x) = 1 – x^2 + x^3$, $a = -1$, and $b = 1$. We can calculate that $f'(x) = 2x – 3x^2$. So, the mean value theorem tells us that there exists a point $c$ in $(-1, 1)$ such that $2c – 3c^2 = \frac{1 – (-1)}{1 – (-1)} = 2$. Solving for $c$, we get $c = \frac{1}{2}$.
Therefore, the value of $x$ in the open interval $(-1, 1)$ for which the mean value theorem is satisfied is $\boxed{\frac{1}{2}}$.
We can also see that the graph of $f$ intersects the line $y = 2$ at $x = \frac{1}{2}$. This is a visual confirmation that the mean value theorem is satisfied at $x = \frac{1}{2}$.