Inverse Function Partner Share - mathcoachblog - Free Printable

Actually, there is no explicit “task” or question stated in the image you uploaded — it simply lists 8 different functions labeled 1A through 4B.

Since no specific instruction (like “find inverses,” “graph,” “solve for x,” “match pairs,” etc.) is given, I’ll assume the most common and logical task when presented with such paired functions:
👉 Find which functions are inverses of each other.

This is a standard Algebra 2 / Precalculus exercise — matching functions with their inverses.

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## ✔ Strategy:
To check if two functions are inverses, we verify that:
> f(g(x)) = x and g(f(x)) = x

If both compositions return `x`, then `f` and `g` are inverses.

We’ll test each A-B pair from the same row (1A & 1B, 2A & 2B, etc.).

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## 🔍 Let’s Test Each Pair:

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✔ Pair 1A & 1B:

- 1A: `f(x) = 2x - 7`
- 1B: `g(x) = (x + 7)/2`

Compute `f(g(x))`:
```
f(g(x)) = f( (x+7)/2 ) = 2*( (x+7)/2 ) - 7 = (x + 7) - 7 = x ✔
```

Compute `g(f(x))`:
```
g(f(x)) = g(2x - 7) = ( (2x - 7) + 7 ) / 2 = (2x)/2 = x ✔
```

✔ 1A and 1B are inverses.

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✘ Pair 2A & 2B:

- 2A: `f(x) = 3x² - 1`
- 2B: `g(x) = √[(x + 1)/3]`

Note: `f(x)` is a quadratic, so it’s not one-to-one over all real numbers → no inverse unless domain restricted.

But even ignoring that, let’s test composition.

Compute `f(g(x))`:
```
f(g(x)) = 3*[√((x+1)/3)]² - 1 = 3 * [(x+1)/3] - 1 = (x+1) - 1 = x ✔
```

Now compute `g(f(x))`:
```
g(f(x)) = √[ ( (3x² - 1) + 1 ) / 3 ] = √[ (3x²)/3 ] = √(x²) = |x| ✘
```

This equals `|x|`, not `x` — so only true if x ≥ 0.

Thus, they are NOT full inverses over all real numbers.

🚫 2A and 2B are NOT inverses (unless domain is restricted to x ≥ 0).

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✔ Pair 3A & 3B:

- 3A: `f(x) = (x - 5)/3`
- 3B: `g(x) = 3x + 5`

Compute `f(g(x))`:
```
f(g(x)) = f(3x + 5) = ( (3x + 5) - 5 ) / 3 = (3x)/3 = x ✔
```

Compute `g(f(x))`:
```
g(f(x)) = g( (x-5)/3 ) = 3*( (x-5)/3 ) + 5 = (x - 5) + 5 = x ✔
```

✔ 3A and 3B are inverses.

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✘ Pair 4A & 4B:

- 4A: `f(x) = 2x³ + 1`
- 4B: `g(x) = ∛[(x - 1)/2]`

Compute `f(g(x))`:
```
f(g(x)) = 2*[∛((x-1)/2)]³ + 1 = 2*[(x-1)/2] + 1 = (x - 1) + 1 = x ✔
```

Compute `g(f(x))`:
```
g(f(x)) = ∛[ ( (2x³ + 1) - 1 ) / 2 ] = ∛[ (2x³)/2 ] = ∛(x³) = x ✔
```

✔ 4A and 4B are inverses!

Wait — this works perfectly!

Cubing and cube root are inverses over all real numbers (unlike square root), so no domain issues.

✔ So 4A and 4B ARE inverses.

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## 🎯 Final Answer:

The pairs of inverse functions are:

- 1A and 1B
- 3A and 3B
- 4A and 4B

2A and 2B are NOT inverses (due to the square root only returning non-negative outputs, while the quadratic can take any real input).

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## 💡 Bonus Tip:
You can also find the inverse of a function algebraically:

1. Replace `f(x)` with `y`.
2. Swap `x` and `y`.
3. Solve for `y`.
4. That’s your inverse.

Example for 1A: `y = 2x - 7` → swap → `x = 2y - 7` → solve → `y = (x + 7)/2` → matches 1B ✔

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## ✔ Summary:

| Pair | Are Inverses? |
|------|---------------|
| 1A & 1B | ✔ Yes |
| 2A & 2B | ✘ No (domain issue) |
| 3A & 3B | ✔ Yes |
| 4A & 4B | ✔ Yes |

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Let me know if you meant a different task — like graphing, evaluating, or solving equations — and I’ll adjust accordingly!