If the functions `f(x)=x^(3)+e^(x//2) " and " g(x)=f^(-1)(x)`, 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 use the relationship between the derivatives of inverse functions.

### Step-by-Step Solution:

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

2. **Find \( f'(x) \)**: We need to differentiate \( f(x) \):
\[ f(x) = x^3 + e^{x/2} \]
Using the power rule and the chain rule:
\[ f'(x) = 3x^2 + \frac{1}{2} e^{x/2} \]

3. **Evaluate \( f(0) \)**: We need to find \( f(0) \) to determine \( g(1) \):
\[ f(0) = 0^3 + e^{0/2} = 1 \]
Thus, \( f(0) = 1 \) implies \( g(1) = 0 \).

4. **Evaluate \( f'(0) \)**: Now we find \( f'(0) \):
\[ f'(0) = 3(0)^2 + \frac{1}{2} e^{0/2} = 0 + \frac{1}{2} = \frac{1}{2} \]

5. **Substitute into the derivative relationship**: We now substitute \( x = 0 \) into the derivative relationship:
\[ g'(f(0)) \cdot f'(0) = 1 \implies g'(1) \cdot f'(0) = 1 \]
Substituting \( f'(0) = \frac{1}{2} \):
\[ g'(1) \cdot \frac{1}{2} = 1 \]

6. **Solve for \( g'(1) \)**:
\[ g'(1) = 1 \cdot 2 = 2 \]

### Final Answer:
Thus, the value of \( g'(1) \) is \( 2 \).