Step-by-step Solution:

1. Understand the relationship between the derivative of a function and its inverse. We are given two functions, $f(x)$ and $g(x)$, where $g(x)$ is the inverse of $f(x)$, i.e., $g(x) = f^{-1}(x)$. The formula for the derivative of an inverse function states that if $y = f(x)$, then $x = g(y)$, and the derivative of $g(y)$ with respect to $y$ is given by: $g'(y) = \frac{1}{f'(x)}$

2. Identify the specific value we need to compute. We are asked to find the value of $g'(1)$. According to the formula, this means we need to find the value of $x$ for which $f(x) = 1$. $g'(1) = \frac{1}{f'(x)} \quad \text{where } f(x) = 1$

3. Find the value of x for which f(x) = 1. The function is given by $f(x) = x^3 + e^{x/2}$. We need to solve the equation: $x^3 + e^{x/2} = 1$ This equation is not easily solvable algebraically. We can try to find a solution by inspection, testing simple values for $x$. Let's try $x=0$: $f(0) = 0^3 + e^{0/2} = 0 + e^0 = 0 + 1 = 1$ So, we found that when $y=1$, the corresponding value of $x$ is $0$.

4. Find the derivative of f(x). The function is $f(x) = x^3 + e^{x/2}$. We differentiate it with respect to $x$ using the power rule and the chain rule. $f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(e^{x/2})$ $f'(x) = 3x^2 + e^{x/2} \cdot \frac{d}{dx}\left(\frac{x}{2}\right)$ $f'(x) = 3x^2 + e^{x/2} \cdot \frac{1}{2}$

5. Evaluate the derivative f'(x) at x = 0. Now we substitute the value $x=0$ (found in step 3) into the expression for $f'(x)$: $f'(0) = 3(0)^2 + \frac{1}{2}e^{0/2} = 3(0) + \frac{1}{2}e^0 = 0 + \frac{1}{2}(1) = \frac{1}{2}$

6. Calculate g'(1) using the inverse function derivative formula. Using the formula from step 1 with $y=1$ and $x=0$: $g'(1) = \frac{1}{f'(0)}$ Substitute the value of $f'(0)$ we found in step 5: $g'(1) = \frac{1}{1/2} = 2$

Conclusion:

The value of $g'(1)$ is 2.