To solve the given problem, we will address both parts step by step.

Part (a): Find an equation for $f^{-1}$, the inverse function.

The given function is:

$f(x) = 6x + 7$

To find the inverse:

1. Replace $f(x)$ with $y$: $y = 6x + 7$

2. Swap $x$ and $y$: $x = 6y + 7$

3. Solve for $y$ to find the inverse: $x - 7 = 6y$ $y = \frac{x - 7}{6}$

Thus, the inverse function is:

$f^{-1}(x) = \frac{x - 7}{6}$

Part (b): Verify that your equation is correct by showing that:

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

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

Substitute $f^{-1}(x) = \frac{x - 7}{6}$ into $f(x)$: $f(f^{-1}(x)) = f\left(\frac{x - 7}{6}\right)$ $f\left(\frac{x - 7}{6}\right) = 6\left(\frac{x - 7}{6}\right) + 7$ Simplify: $6 \cdot \frac{x - 7}{6} + 7 = (x - 7) + 7 = x$ Thus, $f(f^{-1}(x)) = x$.

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

Substitute $f(x) = 6x + 7$ into $f^{-1}(x)$: $f^{-1}(f(x)) = f^{-1}(6x + 7)$ $f^{-1}(6x + 7) = \frac{(6x + 7) - 7}{6}$ Simplify: $\frac{6x + 7 - 7}{6} = \frac{6x}{6} = x$ Thus, $f^{-1}(f(x)) = x$.

Final Answer:

1. The inverse function is: $f^{-1}(x) = \frac{x - 7}{6}$

2. The verification shows: $f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x$

Would you like further clarifications or additional examples?

Related Questions:

1. What are the conditions for a function to have an inverse?
2. Can the inverse of $f(x) = ax + b$ always be expressed in the same form?
3. What happens if a function is not one-to-one? Does it have an inverse?
4. How does the graph of a function compare to the graph of its inverse?
5. Can we verify inverses graphically using reflection across the line $y = x$?

Tip:

When finding the inverse of a function, always verify by substituting into both compositions $f(f^{-1}(x))$ and $f^{-1}(f(x))$ to ensure correctness.