If the function ` f(x)=x^3+e^(x/2)` and `g(x)=f ^(−1)(x)`, then the value of ` g ′ (1)` is

Answer

To find the value of \( g'(1) \) where \( g(x) = f^{-1}(x) \) and \( f(x) = x^3 + e^{x/2} \), we can follow these steps:

### Step 1: Understand the relationship between \( g(x) \) and \( f(x) \)
Since \( g(x) \) is the inverse of \( f(x) \), we have:
\[ g(f(x)) = x \]
Differentiating both sides with respect to \( x \):
\[ g'(f(x)) \cdot f'(x) = 1 \]

### Step 2: Solve for \( g'(f(x)) \)
From the equation above, we can express \( g'(f(x)) \):
\[ g'(f(x)) = \frac{1}{f'(x)} \]

### Step 3: Find \( x \) such that \( f(x) = 1 \)
We need to find the value of \( x \) for which \( f(x) = 1 \):
\[ f(x) = x^3 + e^{x/2} = 1 \]
Testing \( x = 0 \):
\[ f(0) = 0^3 + e^{0/2} = 0 + 1 = 1 \]
Thus, \( f(0) = 1 \) implies \( g(1) = 0 \).

### Step 4: Substitute \( x = 0 \) into \( g'(f(x)) \)
Now, we can substitute \( x = 0 \) into the expression for \( g'(f(x)) \):
\[ g'(1) = g'(f(0)) = \frac{1}{f'(0)} \]

### Step 5: Calculate \( f'(x) \)
Now, we need to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + e^{x/2}) = 3x^2 + \frac{1}{2} e^{x/2} \]
Now, substitute \( x = 0 \):
\[ f'(0) = 3(0)^2 + \frac{1}{2} e^{0/2} = 0 + \frac{1}{2} \cdot 1 = \frac{1}{2} \]

### Step 6: Calculate \( g'(1) \)
Now substituting \( f'(0) \) back into the equation for \( g'(1) \):
\[ g'(1) = \frac{1}{f'(0)} = \frac{1}{\frac{1}{2}} = 2 \]

### Final Answer
Thus, the value of \( g'(1) \) is:
\[ \boxed{2} \]