Question

Determine algebraically whether k(x) and j(x) are inverse functions. True or False.

k(x)=7−x4−2x and j(x)=4x+62x−1

(1 point)
Responses

True
True

False

Answer 1

To determine whether \( k(x) \) and \( j(x) \) are inverse functions, we need to check if the following holds:

1. \( k(j(x)) = x \)
2. \( j(k(x)) = x \)

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**.