If function `f(x)=x^(2)+e^(x//2) " and " g(x)=f^(-1)(x)`, then the value of g'(1) is

Answer

To solve the problem, we need to find the value of \( g'(1) \) where \( g(x) = f^{-1}(x) \) and \( f(x) = x^2 + e^{x/2} \).

### Step-by-step Solution:

1. **Understanding the Functions**:
We have the function \( f(x) = x^2 + e^{x/2} \) and its inverse \( g(x) = f^{-1}(x) \).

2. **Using the Inverse Function Derivative Formula**:
The derivative of the inverse function can be expressed as:
\[ g'(x) = \frac{1}{f'(g(x))} \]
We need to find \( g'(1) \), so we will first find \( g(1) \).

3. **Finding \( x \) such that \( f(x) = 1 \)**:
We need to determine \( x \) for which \( f(x) = 1 \):
\[ f(x) = x^2 + e^{x/2} = 1 \]
Let's check \( x = 0 \):
\[ f(0) = 0^2 + e^{0/2} = 0 + 1 = 1 \]
Therefore, \( g(1) = 0 \).

4. **Calculating \( f'(x) \)**:
Now we need to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^2 + e^{x/2}) = 2x + \frac{1}{2} e^{x/2} \]

5. **Evaluating \( f'(g(1)) \)**:
Since \( g(1) = 0 \), we need to compute \( f'(0) \):
\[ f'(0) = 2(0) + \frac{1}{2} e^{0/2} = 0 + \frac{1}{2} \cdot 1 = \frac{1}{2} \]

6. **Finding \( g'(1) \)**:
Now we can substitute back into the inverse function derivative formula:
\[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)} = \frac{1}{\frac{1}{2}} = 2 \]

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