To find \(g'(x)\), differentiate \(g(x) = 2(x-1)^3 + 8\).

Using the chain rule, \(\frac{d}{dx}[(x-1)^3] = 3(x-1)^2 \cdot 1\).

Thus, \(g'(x) = 2 \cdot 3(x-1)^2 = 6(x-1)^2\).

Since \(g'(x) = 6(x-1)^2 > 0\) for \(x > 1\), \(g(x)\) is strictly increasing.

This implies \(g\) is one-to-one (1:1), so \(g\) has an inverse.