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.