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Let's analyze the given table and the problem step by step.

### Given Information:
- The table provides values for a function \( f(x) \) and its derivative \( f'(x) \) at \( x = 2 \), \( x = 3 \), and \( x = 4 \).
- You are asked to find the equation of the tangent line to the graph of \( g \) at \( x = 2 \), where \( g \) is the inverse function of \( f \).

### Key Concepts:
- If \( g(x) \) is the inverse of \( f(x) \), the relationship between the derivatives of \( f \) and \( g \) at corresponding points is given by:
  \[
  g'(f(a)) = \frac{1}{f'(a)}
  \]
  where \( g(f(a)) = a \).

### Steps:

1. **Identify Corresponding Points**: Since \( g \) is the inverse of \( f \), we have the relationship \( g(f(a)) = a \). To find the tangent at \( x = 2 \) for \( g \), you need to find the value of \( f(x) \) that corresponds to \( g(2) = x \). 

   From the table, \( f(3) = 2 \), meaning \( g(2) = 3 \). So, we are looking for the tangent line to \( g(x) \) at \( x = 2 \), and at this point, \( g(2) = 3 \).

2. **Find the Derivative of \( g \) at \( x = 2 \)**:
   We need \( g'(2) \). The formula for the derivative of the inverse function tells us:
   \[
   g'(2) = \frac{1}{f'(3)}
   \]
   From the table, \( f'(3) = 5 \), so:
   \[
   g'(2) = \frac{1}{5}
   \]

3. **Equation of the Tangent Line**:
   The equation of a tangent line is generally of the form:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency.

   Here, \( m = g'(2) = \frac{1}{5} \), and the point of tangency is \( (2, 3) \), because \( g(2) = 3 \).

   Therefore, the equation of the tangent line is:
   \[
   y - 3 = \frac{1}{5}(x - 2)
   \]
   Simplifying:
   \[
   y = \frac{1}{5}(x - 2) + 3
   \]

### Conclusion:
This matches option **(C)**: \( y = \frac{1}{5}(x - 2) + 3 \).

Would you like any further details or clarifications on this problem?

Here are some related questions for further practice:
1. What is the general relationship between the derivative of a function and its inverse?
2. How do you find the equation of a tangent line for a normal function (not its inverse)?
3. What happens to the slope of the tangent line if \( f'(x) \) is zero at some point?
4. Can you derive the tangent line equation for a different point, say \( x = 3 \)?
5. How does the formula for \( g'(x) \) change if the derivative of \( f(x) \) is negative?

**Tip**: When dealing with inverse functions, always remember that the slope of the tangent line at a point on the inverse function is the reciprocal of the slope at the corresponding point on the original function.

The table above gives values of the differentiable function f and its derivative at selected values of z. If g is the inverse function of f, which of the following is an equation of the line tangent to the graph of g at the point where x = 2?

Learn how to find the equation of the tangent line to the inverse function g(x) at the point where x = 2, using values of f(x) and f'(x). This problem involves applying the inverse function theorem and differentiation concepts. Ideal for students in advanced calculus.

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