Let g(x) be the inverse of f(x) = x3 + 2x + 1. Find g'(4).

We set $f(x) = 4$:

$$x^3 + 2x + 1 = 4$$ $$x^3 + 2x - 3 = 0$$

By inspection, we can test integer values for $x$.
If $x=1$:

$$(1)^3 + 2(1) - 3 = 1 + 2 - 3 = 0$$

So, $x=1$ is the value for which $f(x)=4$. This means $g(4)=1$.

$$f(x) = x^3 + 2x + 1$$ $$f'(x) = \frac{d}{dx}(x^3 + 2x + 1)$$ $$f'(x) = 3x^2 + 2$$

$$f'(1) = 3(1)^2 + 2$$ $$f'(1) = 3 + 2$$ $$f'(1) = 5$$

$$g'(4) = \frac{1}{f'(1)}$$ $$g'(4) = \frac{1}{5}$$

The final answer is $\frac{1}{5}$.