Derivative of Inverse Function at Specified Point

To find the value of $g'(1)$ for the inverse function $g(x) = f^{-1}(x)$, we use the relationship between the derivative of a function and the derivative of its inverse:

$g(f(x)) = x$

Differentiating both sides with respect to $x$ using the chain rule gives:
$g'(f(x)) \cdot f'(x) = 1 \implies g'(f(x)) = \frac{1}{f'(x)}$

We need to evaluate $g'(1)$, which means we must find the value of $x$ such that:
$f(x) = 1$

Given:
$f(x) = x^3 + e^{\frac{x}{2}}$

Setting $f(x) = 1$:
$x^3 + e^{\frac{x}{2}} = 1$

By inspection, when $x = 0$:
$f(0) = 0^3 + e^0 = 1$

Since the derivative of $f(x)$ is:
$f'(x) = 3x^2 + \frac{1}{2}e^{\frac{x}{2}} > 0 \quad \text{for all } x \in \mathbb{R}$

$f(x)$ is strictly increasing, which guarantees that $x = 0$ is the unique solution to $f(x) = 1$.

Now, we evaluate $f'(x)$ at $x = 0$:
$f'(0) = 3(0)^2 + \frac{1}{2}e^0 = \frac{1}{2}$

Finally, we calculate $g'(1)$:
$g'(1) = \frac{1}{f'(0)} = \frac{1}{\frac{1}{2}} = 2$