1. **State the problem:** We are given a one-to-one function $$f(x) = \frac{5x - 7}{7x + 5}$$ and need to find its inverse function $$f^{-1}(x)$$. 2. **Formula and rules:** To find the inverse function, we swap $$x$$ and $$y$$ in the equation and solve for $$y$$. 3. **Step-by-step solution:** Let $$y = f(x) = \frac{5x - 7}{7x + 5}$$. Swap $$x$$ and $$y$$: $$x = \frac{5y - 7}{7y + 5}$$ Multiply both sides by $$7y + 5$$ to clear the denominator: $$x(7y + 5) = 5y - 7$$ Distribute $$x$$: $$7xy + 5x = 5y - 7$$ Group terms with $$y$$ on one side: $$7xy - 5y = -7 - 5x$$ Factor out $$y$$: $$y(7x - 5) = -7 - 5x$$ Divide both sides by $$7x - 5$$: $$y = \frac{-7 - 5x}{7x - 5}$$ Show cancellation step (if any): No common factors to cancel here. 4. **Domain and range:** - The domain of $$f$$ is all real numbers except where the denominator is zero: $$7x + 5 = 0 \Rightarrow x = -\frac{5}{7}$$ - The range of $$f$$ is all real numbers except where the denominator of $$f^{-1}$$ is zero: $$7x - 5 = 0 \Rightarrow x = \frac{5}{7}$$ Thus: - Domain of $$f^{-1}$$ is $$\mathbb{R} \setminus \left\{ \frac{5}{7} \right\}$$ - Range of $$f^{-1}$$ is $$\mathbb{R} \setminus \left\{ -\frac{5}{7} \right\}$$ In interval notation: - Domain of $$f^{-1}$$: $$(-\infty, \frac{5}{7}) \cup (\frac{5}{7}, \infty)$$ - Range of $$f^{-1}$$: $$(-\infty, -\frac{5}{7}) \cup (-\frac{5}{7}, \infty)$$