Instructions. Use algebraic techniques to solve problems, unless otherwise instructed. Shorw the calculations leading to your answers (except basic arithmetic), when there is work to show. Simplify and circle your final answers. 1. (12 pts.) (a) Find the inverse function of \( f \), where \( f(x)=\frac{5 x}{2 x-7} \), (b) Use function composition to prove that the two functions are inverses. (c) Find the domain and range of each function.

1. **Finding the inverse function** We start with \[ y=\frac{5x}{2x-7}. \] To find the inverse, solve for \( x \) in terms of \( y \): Multiply both sides by \( (2x-7) \): \[ y(2x-7)=5x. \] Distribute \( y \) on the left-hand side: \[ 2yx-7y=5x. \] Rearrange the equation to isolate terms containing \( x \): \[ 2yx-5x=7y. \] Factor \( x \) out of the left-hand side: \[ x(2y-5)=7y. \] Solve for \( x \): \[ x=\frac{7y}{2y-5}. \] Now, replace \( y \) with \( x \) to denote the inverse function: \[ f^{-1}(x)=\frac{7x}{2x-5}. \] Circle the final answer: \(\boxed{f^{-1}(x)=\frac{7x}{2x-5}}\). 2. **Verifying with function composition** - Compute \( f(f^{-1}(x)) \): Substitute \( f^{-1}(x)=\frac{7x}{2x-5} \) into \( f \): \[ f(f^{-1}(x)) = \frac{5\left(\frac{7x}{2x-5}\right)}{2\left(\frac{7x}{2x-5}\right)-7}. \] Simplify the numerator: \[ 5\left(\frac{7x}{2x-5}\right)=\frac{35x}{2x-5}. \] Simplify the denominator: \[ 2\left(\frac{7x}{2x-5}\right)-7 = \frac{14x}{2x-5}-7. \] Express \( 7 \) with a common denominator: \[ \frac{14x}{2x-5}-\frac{7(2x-5)}{2x-5} = \frac{14x-14x+35}{2x-5} = \frac{35}{2x-5}. \] Now, the expression becomes: \[ f(f^{-1}(x))=\frac{\frac{35x}{2x-5}}{\frac{35}{2x-5}}=\frac{35x}{2x-5}\cdot\frac{2x-5}{35}=x. \] - Compute \( f^{-1}(f(x)) \): Substitute \( f(x)=\frac{5x}{2x-7} \) into \( f^{-1} \): \[ f^{-1}(f(x))=\frac{7\left(\frac{5x}{2x-7}\right)}{2\left(\frac{5x}{2x-7}\right)-5}. \] Simplify the numerator: \[ 7\left(\frac{5x}{2x-7}\right)=\frac{35x}{2x-7}. \] Simplify the denominator: \[ 2\left(\frac{5x}{2x-7}\right)-5=\frac{10x}{2x-7}-5. \] Express \( 5 \) with a common denominator: \[ \frac{10x}{2x-7}-\frac{5(2x-7)}{2x-7}=\frac{10x-10x+35}{2x-7}=\frac{35}{2x-7}. \] Now, the expression becomes: \[ f^{-1}(f(x))=\frac{\frac{35x}{2x-7}}{\frac{35}{2x-7}}=\frac{35x}{2x-7}\cdot\frac{2x-7}{35}=x. \] Since both compositions return \( x \), the functions are inverse to each other. 3. **Finding the domain and range** - **Original function \( f(x)=\frac{5x}{2x-7} \):** The denominator cannot be zero: \[ 2x-7\neq 0 \quad \Rightarrow \quad x\neq \frac{7}{2}. \] So, the domain of \( f \) is: \[ \{x\in\mathbb{R}: x\neq \frac{7}{2}\}. \] To find the range, solve for \( y \) in terms of \( x \): \[ y=\frac{5x}{2x-7} \quad \Rightarrow \quad x=\frac{7y}{2y-5} \quad \text{(as obtained earlier)}. \] The expression for \( x \) is defined if the denominator is not zero: \[ 2y-5\neq 0 \quad \Rightarrow \quad y\neq \frac{5}{2}. \] Hence, the range of \( f \) is: \[ \{y\in\mathbb{R}: y\neq \frac{5}{2}\}. \] - **Inverse function \( f^{-1}(x)=\frac{7x}{2x-5} \):** Similarly, the denominator cannot be zero: \[ 2x-5\neq 0 \quad \Rightarrow \quad x\neq \frac{5}{2}. \] So, the domain of \( f^{-1} \) is: \[ \{x\in\mathbb{R}: x\neq \frac{5}{2}\}. \] And the range of \( f^{-1} \), being the domain of \( f \), is: \[ \{y\in\mathbb{R}: y\neq \frac{7}{2}\}. \] (Note that since \( f \) and \( f^{-1} \) are inverses, the domain of one is the range of the other.) Final answers: - \(\boxed{f^{-1}(x)=\frac{7x}{2x-5}}\). - Verification: \( f(f^{-1}(x))=x \) and \( f^{-1}(f(x))=x \). - Domain of \( f(x) \): \(\{x\in\mathbb{R}: x\neq \frac{7}{2}\}\). - Range of \( f(x) \): \(\{y\in\mathbb{R}: y\neq \frac{5}{2}\}\). - Domain of \( f^{-1}(x) \): \(\{x\in\mathbb{R}: x\neq \frac{5}{2}\}\). - Range of \( f^{-1}(x) \): \(\{y\in\mathbb{R}: y\neq \frac{7}{2}\}\).