Inverse Function Partner Share - mathcoachblog - Free Printable

It looks like you've uploaded a list of functions labeled 1A, 1B, 2A, 2B, etc., but no specific task or question was stated. These are just eight different functions:

- Linear: $ f(x) = 2x - 7 $, $ f(x) = \frac{x+7}{2} $, $ f(x) = \frac{x-5}{3} $, $ f(x) = 3x + 5 $
- Quadratic: $ f(x) = 3x^2 - 1 $
- Cubic: $ f(x) = 2x^3 + 1 $
- Square root: $ f(x) = \sqrt{\frac{x+1}{3}} $
- Cube root: $ f(x) = \sqrt[3]{\frac{x-1}{2}} $

Since the problem is not specified, here are common tasks that might be associated with such a list — and I’ll solve them for each function where applicable.

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Possible Task: Find the inverse function $ f^{-1}(x) $ for each function.

Let’s assume the task is to find the inverse of each function, if it exists.

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1A: $ f(x) = 2x - 7 $

Step 1: Replace $ f(x) $ with $ y $:
$ y = 2x - 7 $

Step 2: Solve for $ x $:
$ y + 7 = 2x $
$ x = \frac{y + 7}{2} $

Step 3: Swap $ x $ and $ y $:
$ y = \frac{x + 7}{2} $

✔ So, $ f^{-1}(x) = \frac{x + 7}{2} $

> This matches 1B, so 1A and 1B are inverses.

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

We already found its inverse above.

Step 1: $ y = \frac{x + 7}{2} $

Step 2: Multiply both sides by 2:
$ 2y = x + 7 $ → $ x = 2y - 7 $

Step 3: Swap: $ y = 2x - 7 $

✔ So, $ f^{-1}(x) = 2x - 7 $ → matches 1A

So again, 1A and 1B are inverses.

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2A: $ f(x) = 3x^2 - 1 $

This is a quadratic function (parabola), which is not one-to-one over all real numbers, so it does not have an inverse unless we restrict the domain.

But let's try to find an inverse anyway.

Step 1: $ y = 3x^2 - 1 $

Step 2: $ y + 1 = 3x^2 $ → $ x^2 = \frac{y + 1}{3} $

Step 3: $ x = \pm \sqrt{\frac{y + 1}{3}} $

So, no unique inverse unless we restrict domain.

✔ So, inverse does not exist globally.

But notice: 2B is $ f(x) = \sqrt{\frac{x+1}{3}} $, which is half of this inverse.

So, if we restrict 2A to $ x \geq 0 $, then $ f^{-1}(x) = \sqrt{\frac{x+1}{3}} $ → matches 2B

Thus:
- 2A and 2B are inverses if domain of 2A is restricted to $ x \geq 0 $

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

This is defined only when $ \frac{x+1}{3} \geq 0 $ → $ x \geq -1 $

Let’s find its inverse.

Step 1: $ y = \sqrt{\frac{x+1}{3}} $

Step 2: Square both sides: $ y^2 = \frac{x+1}{3} $

→ $ 3y^2 = x + 1 $ → $ x = 3y^2 - 1 $

Swap: $ y = 3x^2 - 1 $

✔ So $ f^{-1}(x) = 3x^2 - 1 $, which is 2A

But note: since $ f(x) = \sqrt{\cdots} \geq 0 $, the inverse must be defined for $ x \geq 0 $

So again, 2A and 2B are inverses with appropriate domain restrictions.

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

Step 1: $ y = \frac{x - 5}{3} $

Step 2: $ 3y = x - 5 $ → $ x = 3y + 5 $

Step 3: Swap: $ y = 3x + 5 $

✔ So $ f^{-1}(x) = 3x + 5 $ → matches 3B

So 3A and 3B are inverses

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3B: $ f(x) = 3x + 5 $

We already did this.

Step 1: $ y = 3x + 5 $

Step 2: $ y - 5 = 3x $ → $ x = \frac{y - 5}{3} $

Step 3: Swap: $ y = \frac{x - 5}{3} $

✔ So $ f^{-1}(x) = \frac{x - 5}{3} $ → matches 3A

So again, 3A and 3B are inverses

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4A: $ f(x) = 2x^3 + 1 $

Step 1: $ y = 2x^3 + 1 $

Step 2: $ y - 1 = 2x^3 $ → $ x^3 = \frac{y - 1}{2} $

Step 3: $ x = \sqrt[3]{\frac{y - 1}{2}} $

Swap: $ y = \sqrt[3]{\frac{x - 1}{2}} $

✔ So $ f^{-1}(x) = \sqrt[3]{\frac{x - 1}{2}} $ → matches 4B

So 4A and 4B are inverses

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

Cube both sides: $ y^3 = \frac{x - 1}{2} $

→ $ 2y^3 = x - 1 $ → $ x = 2y^3 + 1 $

Swap: $ y = 2x^3 + 1 $

✔ So $ f^{-1}(x) = 2x^3 + 1 $ → matches 4A

So 4A and 4B are inverses

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✔ Final Summary:

| Pair | Are they inverses? | Notes |
|------|--------------------|-------|
| 1A & 1B | Yes | Linear functions, inverses |
| 2A & 2B | Yes | With domain restriction $ x \geq 0 $ for 2A |
| 3A & 3B | Yes | Linear, inverses |
| 4A & 4B | Yes | Cubic and cube root, inverses |

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🟩 Conclusion:

If the task was to find which pairs are inverses, then:

> The inverse pairs are:
> - 1A ↔ 1B
> - 2A ↔ 2B (with domain restriction)
> - 3A ↔ 3B
> - 4A ↔ 4B

If your original task was something else (like finding domain, range, evaluating at a point, graphing, etc.), please clarify, and I can adjust accordingly.

But based on the structure, finding inverses is the most likely intended task.