If the function f(x) = x³ + ex/2 and g (x) = fn(x), then the value of g'(1) is

The correct answer is 2.

Step 1: Understand the Inverse Function Relationship
Let the given function be:
$$f(x)=x^3+e^{x/2}$$
We are given that $g(x)=f^{-1}(x)$, which represents the inverse function of $f(x)$.
Using the derivative rule for inverse functions, we have:
$$g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$$
Substituting $x=1$ gives:
$$g^{\prime }(1)=\frac{1}{f^{\prime }(g(1))}$$

Step 2: Find the value of $g(1)$
Let $g(1)=y$. This means that $f(y)=1$.
Substitute $y$ into the function $f(x)$:
$$y^3+e^{y/2}=1$$
By observation, if we set $y=0$:
$$0^3+e^{0/2}=0+1=1$$
Since the function $f(x)$ is strictly increasing, $y=0$ is the unique real solution. Therefore, $g(1)=0$.

Step 3: Differentiate $f(x)$
Differentiating $f(x)=x^3+e^{x/2}$ with respect to $x$ using the power rule and chain rule:
$$f^{\prime }(x)=3x^2+\frac{1}{2}$$

Step 4: Calculate $f^{\prime }(g(1))$
Substitute $g(1)=0$ into the derivative expression:
$$f^{\prime }(0)=3(0{)}^2+\frac{1}{2}{e}^{0/2}=0+\frac{1}{2}(1)=\frac{1}{2}$$

Step 5: Compute $g^{\prime }(1)$
Using the inverse derivative formula:
$$g^{\prime }(1)=\frac{1}{f^{\prime }(0)}=\frac{1}{1/2}=2$$