1. **Problem statement:** Given the function $f(x) = \frac{2x + 2}{x - 1}$ with $x \neq 1$, find (1) $f^{-1}(3)$ and (2) express $f^{-1}(p) = kp + d$ and determine $k$ and $d$. 2. **Find the inverse function $f^{-1}(x)$:** Start by setting $y = \frac{2x + 2}{x - 1}$. 3. Swap $x$ and $y$ to find the inverse: $$x = \frac{2y + 2}{y - 1}$$ 4. Multiply both sides by $(y - 1)$: $$x(y - 1) = 2y + 2$$ 5. Distribute $x$: $$xy - x = 2y + 2$$ 6. Group terms with $y$ on one side: $$xy - 2y = x + 2$$ 7. Factor out $y$: $$y(x - 2) = x + 2$$ 8. Solve for $y$: $$y = \frac{x + 2}{x - 2}$$ So the inverse function is: $$f^{-1}(x) = \frac{x + 2}{x - 2}$$ 9. **Find $f^{-1}(3)$:** Substitute $x = 3$: $$f^{-1}(3) = \frac{3 + 2}{3 - 2} = \frac{5}{1} = 5$$ 10. **Express $f^{-1}(p)$ in the form $kp + d$:** We have: $$f^{-1}(p) = \frac{p + 2}{p - 2}$$ Rewrite as: $$f^{-1}(p) = \frac{p - 2 + 4}{p - 2} = 1 + \frac{4}{p - 2}$$ This is not a linear function of the form $kp + d$ because of the denominator $p - 2$. 11. **Check if $f^{-1}(p)$ can be linear:** Since $f^{-1}(p)$ is a rational function, it cannot be expressed exactly as $kp + d$ for all $p$. 12. **Alternative interpretation:** If the problem means to find constants $k$ and $d$ such that $f^{-1}(p) = kp + d$ holds approximately or for some domain, it is not possible exactly. **Final answers:** - $f^{-1}(3) = 5$ - $f^{-1}(p)$ cannot be expressed exactly as $kp + d$; it is $\frac{p + 2}{p - 2}$.