Let `f(x) = x + cos x + 2 and g(x)` be the inverse function of f(x), then g'(3) equals to ........ .

Answer

To solve the problem, we need to find \( g'(3) \) where \( g(x) \) is the inverse function of \( f(x) = x + \cos x + 2 \).

### Step-by-Step Solution:

1. **Understanding the Inverse Function**:
Since \( g(x) \) is the inverse of \( f(x) \), we have:
\[ g(f(x)) = x \]
Differentiating both sides with respect to \( x \) gives:
\[ g'(f(x)) \cdot f'(x) = 1 \]
Therefore, we can express \( g'(x) \) in terms of \( f'(x) \):
\[ g'(x) = \frac{1}{f'(g(x))} \]

2. **Finding \( g'(3) \)**:
To find \( g'(3) \), we first need to find \( x \) such that \( f(x) = 3 \):
\[ f(x) = x + \cos x + 2 \]
Setting this equal to 3:
\[ x + \cos x + 2 = 3 \]
Simplifying gives:
\[ x + \cos x = 1 \]

3. **Finding \( x \)**:
We can test \( x = 0 \):
\[ 0 + \cos(0) = 0 + 1 = 1 \]
Thus, \( f(0) = 3 \). Therefore, \( g(3) = 0 \).

4. **Finding the Derivative \( f'(x) \)**:
Now we need to find \( f'(x) \):
\[ f'(x) = 1 - \sin x \]

5. **Evaluating \( f'(g(3)) \)**:
Since \( g(3) = 0 \), we need to evaluate \( f'(0) \):
\[ f'(0) = 1 - \sin(0) = 1 - 0 = 1 \]

6. **Final Calculation of \( g'(3) \)**:
Now substituting back into our expression for \( g'(3) \):
\[ g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(0)} = \frac{1}{1} = 1 \]

### Conclusion:
Thus, the value of \( g'(3) \) is:
\[ \boxed{1} \]