\( 1 \leftarrow \) 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.)

We start with the function   f(x) = (3x) / (7x – 1). Our goals are:  (a) Find the inverse function f⁻¹(x) and check it.  (b) Find the domains and ranges of both f and f⁻¹. ──────────────────────── Step (a): Find f⁻¹(x) 1. Write the equation with y in place of f(x):   y = (3x) / (7x – 1). 2. To find the inverse, swap the roles of x and y:   x = (3y) / (7y – 1). 3. Solve for y:   Multiply both sides by (7y – 1) to clear the denominator:     x(7y – 1) = 3y   Expand the left side:     7xy – x = 3y   Gather terms containing y on one side:     7xy – 3y = x   Factor y out:     y(7x – 3) = x   Solve for y by dividing both sides by (7x – 3):     y = x / (7x – 3). Thus, the inverse function is   f⁻¹(x) = x / (7x – 3). 4. Check the result by computing f(f⁻¹(x)):   Substitute x / (7x – 3) into the original function f:     f(f⁻¹(x)) = f(x / (7x – 3))        = [3 · (x / (7x – 3))] / [7 · (x / (7x – 3)) – 1].   Simplify the numerator:     = 3x / (7x – 3).   Simplify the denominator:     7 · (x / (7x – 3)) – 1 = (7x/(7x – 3)) – 1        = [7x – (7x – 3)] / (7x – 3)        = 3 / (7x – 3).   Now, f(f⁻¹(x)) becomes:     = [3x/(7x – 3)] / [3/(7x – 3)]     = x. This confirms that f⁻¹(x) is the correct inverse. ──────────────────────── Step (b): Domain and Range 1. Domain of f:   f(x) = 3x/(7x – 1) is undefined when its denominator equals zero.   Set 7x – 1 = 0:     7x = 1 ⟹ x = 1/7.   So, the domain of f is:     Domain(f) = { all real numbers } \ {1/7}. 2. Range of f:   Since f is one-to-one, its range is the set of all values that f(x) can take.   Note the horizontal asymptote: When x is very large in magnitude, f(x) ≈ (3x)/(7x) = 3/7.   We check if f(x) ever equals 3/7:     Set (3x)/(7x – 1) = 3/7 and solve:      Multiply both sides by 7(7x – 1):       21x = 3(7x – 1)       21x = 21x – 3       Subtract 21x from both sides:       0 = –3, which is impossible.   Thus, f(x) can never be 3/7.   So, the range of f is:     Range(f) = { all real numbers } \ {3/7}. 3. For the inverse function f⁻¹(x):   Recall that the domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.   So,     Domain(f⁻¹) = Range(f) = { all real numbers } \ {3/7}.     Range(f⁻¹) = Domain(f) = { all real numbers } \ {1/7}. ──────────────────────── Final Answers: (a) f⁻¹(x) = x / (7x – 3). (b) • Domain of f: all real numbers except 1/7; Range of f: all real numbers except 3/7.      Domain of f⁻¹: all real numbers except 3/7; Range of f⁻¹: all real numbers except 1/7.