Inverse Function Partner Share - mathcoachblog - Free Printable

To solve the problem, we need to understand the task. The image presents a list of functions categorized into two groups: 1A, 2A, 3A, 4A and 1B, 2B, 3B, 4B. Each function in group A is paired with a function in group B. The task likely involves identifying a relationship or transformation between the functions in each pair.

Let's analyze each pair step by step:

Pair 1: \( f(x) = 2x - 7 \) (1A) and \( f(x) = \frac{x + 7}{2} \) (1B)

- 1A: \( f(x) = 2x - 7 \)
- 1B: \( f(x) = \frac{x + 7}{2} \)

To find the relationship, let's try to express one function in terms of the other:
1. Start with \( f(x) = 2x - 7 \).
2. Solve for \( x \) in terms of \( f(x) \):
\[
y = 2x - 7 \implies y + 7 = 2x \implies x = \frac{y + 7}{2}
\]
3. Notice that \( \frac{x + 7}{2} \) is the inverse function of \( 2x - 7 \).

Thus, the relationship is that 1B is the inverse of 1A.

Pair 2: \( f(x) = 3x^2 - 1 \) (2A) and \( f(x) = \sqrt{\frac{x + 1}{3}} \) (2B)

- 2A: \( f(x) = 3x^2 - 1 \)
- 2B: \( f(x) = \sqrt{\frac{x + 1}{3}} \)

To find the relationship, let's try to express one function in terms of the other:
1. Start with \( f(x) = 3x^2 - 1 \).
2. Solve for \( x \) in terms of \( f(x) \):
\[
y = 3x^2 - 1 \implies y + 1 = 3x^2 \implies x^2 = \frac{y + 1}{3} \implies x = \pm \sqrt{\frac{y + 1}{3}}
\]
3. Notice that \( \sqrt{\frac{x + 1}{3}} \) is the positive branch of the inverse function of \( 3x^2 - 1 \).

Thus, the relationship is that 2B is the positive branch of the inverse of 2A.

Pair 3: \( f(x) = \frac{x - 5}{3} \) (3A) and \( f(x) = 3x + 5 \) (3B)

- 3A: \( f(x) = \frac{x - 5}{3} \)
- 3B: \( f(x) = 3x + 5 \)

To find the relationship, let's try to express one function in terms of the other:
1. Start with \( f(x) = \frac{x - 5}{3} \).
2. Solve for \( x \) in terms of \( f(x) \):
\[
y = \frac{x - 5}{3} \implies 3y = x - 5 \implies x = 3y + 5
\]
3. Notice that \( 3x + 5 \) is the inverse function of \( \frac{x - 5}{3} \).

Thus, the relationship is that 3B is the inverse of 3A.

Pair 4: \( f(x) = 2x^3 + 1 \) (4A) and \( f(x) = \sqrt[3]{\frac{x - 1}{2}} \) (4B)

- 4A: \( f(x) = 2x^3 + 1 \)
- 4B: \( f(x) = \sqrt[3]{\frac{x - 1}{2}} \)

To find the relationship, let's try to express one function in terms of the other:
1. Start with \( f(x) = 2x^3 + 1 \).
2. Solve for \( x \) in terms of \( f(x) \):
\[
y = 2x^3 + 1 \implies y - 1 = 2x^3 \implies x^3 = \frac{y - 1}{2} \implies x = \sqrt[3]{\frac{y - 1}{2}}
\]
3. Notice that \( \sqrt[3]{\frac{x - 1}{2}} \) is the inverse function of \( 2x^3 + 1 \).

Thus, the relationship is that 4B is the inverse of 4A.

Final Answer

The relationships between the pairs are as follows:
1. 1B is the inverse of 1A
2. 2B is the positive branch of the inverse of 2A
3. 3B is the inverse of 3A
4. 4B is the inverse of 4A

\[
\boxed{\text{Each B function is the inverse of its corresponding A function.}}
\]