Let's work through the problem step by step.

a. Find the equation for the inverse function $f^{-1}(x)$

We are given that $f(x) = 6x + 5$. To find the inverse, follow these steps:

1. Replace $f(x)$ with $y$: $y = 6x + 5$
2. Swap $x$ and $y$ to solve for the inverse: $x = 6y + 5$
3. Solve for $y$ (this will be $f^{-1}(x)$): $x - 5 = 6y$ $y = \frac{x - 5}{6}$ So, the inverse function is: $f^{-1}(x) = \frac{x - 5}{6}$

b. Verify the inverse function

We need to verify that:

1. $f(f^{-1}(x)) = x$
2. $f^{-1}(f(x)) = x$

1. Verify $f(f^{-1}(x)) = x$

Substitute $f^{-1}(x) = \frac{x - 5}{6}$ into $f(x)$: $f(f^{-1}(x)) = 6\left( \frac{x - 5}{6} \right) + 5$ Simplify: $= (x - 5) + 5 = x$ So, $f(f^{-1}(x)) = x$.

2. Verify $f^{-1}(f(x)) = x$

Substitute $f(x) = 6x + 5$ into $f^{-1}(x)$: $f^{-1}(f(x)) = \frac{(6x + 5) - 5}{6}$ Simplify: $= \frac{6x}{6} = x$ So, $f^{-1}(f(x)) = x$.

Both conditions are satisfied, confirming that the inverse function $f^{-1}(x) = \frac{x - 5}{6}$ is correct.

Would you like further details or have any questions?

Here are 5 related questions:

1. What is the graphical interpretation of the inverse of a function?
2. How do you determine if a function is one-to-one without graphing it?
3. How does the horizontal line test relate to one-to-one functions?
4. What are the conditions for a function to have an inverse?
5. How do we handle finding the inverse of a quadratic function?

Tip: To quickly check if a function is one-to-one, see if it passes the horizontal line test: no horizontal line should intersect the graph more than once.