\( 1 \leftarrow \) The function \( f(x)=\frac{3 x}{7 x-1} \) is one-to-one (a) Find its inverse function \( f^{-1} \) and check your answer. (b) Find the domain and the range of \( f \) and \( f^{-1} \). (a) \( f^{-1}(x)=\square \) (Simplify your answer.)

To find the inverse function \( f^{-1}(x) \) of \( f(x) = \frac{3x}{7x - 1} \), we start by letting \( y = f(x) \): \[ y = \frac{3x}{7x - 1} \] Now, we solve for \( x \) in terms of \( y \). First, we'll multiply both sides by \( 7x - 1 \): \[ y(7x - 1) = 3x \] Expanding gives us: \[ 7xy - y = 3x \] We can rearrange this to isolate \( x \): \[ 7xy - 3x = y \] Factoring \( x \) out of the left side: \[ x(7y - 3) = y \] Now, divide both sides by \( 7y - 3 \): \[ x = \frac{y}{7y - 3} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{x}{7x - 3} \] To check our work, we can verify if \( f(f^{-1}(x)) = x \): \[ f\left(f^{-1}(x)\right) = f\left(\frac{x}{7x - 3}\right) = \frac{3\left(\frac{x}{7x - 3}\right)}{7\left(\frac{x}{7x - 3}\right) - 1} \] Calculating the denominator: \[ = 7\left(\frac{x}{7x - 3}\right) - 1 = \frac{7x}{7x - 3} - 1 = \frac{7x - (7x - 3)}{7x - 3} = \frac{3}{7x - 3} \] So: \[ f(f^{-1}(x)) = \frac{3\left(\frac{x}{7x - 3}\right)}{\frac{3}{7x - 3}} = x \] Thus, our inverse is verified. For the domain and range: The domain of \( f \) is \( x \neq \frac{1}{7} \) because it's undefined there (as the denominator cannot be zero). The range of \( f \) is all real numbers except \( \frac{3}{7} \) because the output cannot equal \( \frac{3}{7} \) (check by rearranging the equation). Conversely, \( f^{-1} \) has the domain of \( x \neq \frac{3}{7} \) and the range includes all real numbers except \( \frac{1}{7} \). So in summary: (a) \( f^{-1}(x) = \frac{x}{7x - 3} \) (b) Domain of \( f \): \( x \in \mathbb{R}, x \neq \frac{1}{7} \); Range of \( f \): \( y \in \mathbb{R}, y \neq \frac{3}{7} \); Domain of \( f^{-1} \): \( x \in \mathbb{R}, x \neq \frac{3}{7} \); Range of \( f^{-1} \): \( y \in \mathbb{R}, y \neq \frac{1}{7} \).