To determine whether \( k(x) \) and \( j(x) \) are inverse functions, we need to check if the following holds:
- \( k(j(x)) = x \)
- \( j(k(x)) = x \)
Let's compute \( k(j(x)) \) and \( j(k(x)) \).
First, substitute \( j(x) \) into \( k(x) \):
Given: \[ k(x) = 7 - \frac{x}{4 - 2x} \] and \[ j(x) = \frac{4x + 6}{2x - 1} \]
We will find \( k(j(x)) \): \[ k(j(x)) = k\left(\frac{4x + 6}{2x - 1}\right) = 7 - \frac{\frac{4x + 6}{2x - 1}}{4 - 2\cdot\frac{4x + 6}{2x - 1}} \]
Next, compute the denominator: \[ 4 - 2\cdot\frac{4x + 6}{2x - 1} = 4 - \frac{8x + 12}{2x - 1} = \frac{(4(2x - 1) - (8x + 12))}{2x - 1} = \frac{(8x - 4 - 8x - 12)}{2x - 1} = \frac{-16}{2x - 1} \]
Now, substitute back into \( k(j(x)) \): \[ k\left(\frac{4x + 6}{2x - 1}\right) = 7 - \frac{\frac{4x + 6}{2x - 1}}{\frac{-16}{2x - 1}} = 7 + \frac{4x + 6}{16} = 7 + \frac{1}{4}(4x + 6) \] \[ = 7 + x + \frac{3}{2} = x + \frac{17}{2} \]
Since \( k(j(x)) \neq x \), the functions are not inverses based on this computation.
Now, let's compute \( j(k(x)) \): \[ j(k(x)) = j\left(7 - \frac{x}{4 - 2x}\right) = \frac{4(7 - \frac{x}{4 - 2x}) + 6}{2(7 - \frac{x}{4 - 2x}) - 1} \] This computation is lengthy, but given that the first computation already showed that they are not equal when substituting, it’s safe to conclude:
The functions \( k(x) \) and \( j(x) \) are not inverse functions.
Therefore, the answer is False.