Equations with Composite Functions (Worksheet with FULL solutions ... - Free Printable

You’ve uploaded a worksheet titled “FUNCTIONS: EQUATIONS THAT CONTAIN COMPOUND FUNCTIONS” from Maths4Everyone.com, and you’re asking to solve the problem and explain the solution.

Since you didn’t specify which cell (A1, B2, etc.) you want solved, I’ll solve one representative problem from each row — specifically:

- A1: Solve `gf(x) = 21`
- B1: Solve `gf(x) = g(x)`
- C1: Solve `fg(x) = g⁻¹(x)`
- D1: Solve `fg(x) = f⁻¹(x)`

This will give you a comprehensive understanding of how to approach all types of problems on this sheet.

---

## ✔ A1: Solve `gf(x) = 21`

Given:
- `f(x) = 2x - 3`
- `g(x) = x² - 4`

We need to find `gf(x)` — that means `g(f(x))`.

Step 1: Compute `gf(x) = g(f(x))`

Substitute `f(x)` into `g`:

```
gf(x) = g(2x - 3)
= (2x - 3)² - 4
= (4x² - 12x + 9) - 4
= 4x² - 12x + 5
```

Step 2: Set equal to 21

```
4x² - 12x + 5 = 21
→ 4x² - 12x - 16 = 0
```

Divide entire equation by 4:

```
x² - 3x - 4 = 0
```

Factor:

```
(x - 4)(x + 1) = 0
→ x = 4 or x = -1
```

✔ Answer: x = 4 or x = -1

---

## ✔ B1: Solve `gf(x) = g(x)`

Given:
- `f(x) = x + 2`
- `g(x) = x³ + 3`

Find `gf(x) = g(f(x)) = g(x+2) = (x+2)³ + 3`

Set equal to `g(x) = x³ + 3`

Step 1: Write equation

```
(x + 2)³ + 3 = x³ + 3
```

Subtract 3 from both sides:

```
(x + 2)³ = x³
```

Expand left side:

```
x³ + 6x² + 12x + 8 = x³
```

Subtract `x³` from both sides:

```
6x² + 12x + 8 = 0
```

Divide by 2:

```
3x² + 6x + 4 = 0
```

Use quadratic formula:

```
x = [-6 ± √(36 - 48)] / 6
= [-6 ± √(-12)] / 6
= [-6 ± 2i√3] / 6
= [-3 ± i√3] / 3
```

✔ Answer: No real solutions. Complex solutions: x = (-3 ± i√3)/3

*(Note: Depending on level, you may be expected to say “no real solution”)*

---

## ✔ C1: Solve `fg(x) = g⁻¹(x)`

Given:
- `f(x) = x²`
- `g(x) = x - 3`

First, find `fg(x) = f(g(x)) = f(x - 3) = (x - 3)²`

Now, find `g⁻¹(x)` — inverse of `g(x) = x - 3`

To find inverse:

Let `y = x - 3`, swap x and y: `x = y - 3 → y = x + 3`

So `g⁻¹(x) = x + 3`

Step 2: Set equal

```
(x - 3)² = x + 3
```

Expand left:

```
x² - 6x + 9 = x + 3
→ x² - 7x + 6 = 0
```

Factor:

```
(x - 6)(x - 1) = 0
→ x = 6 or x = 1
```

✔ Answer: x = 6 or x = 1

---

## ✔ D1: Solve `fg(x) = f⁻¹(x)`

Given:
- `f(x) = x/2 + 3`
- `g(x) = 14/(2x - 3)`

Step 1: Find `fg(x) = f(g(x))`

```
fg(x) = f(14/(2x - 3)) = [14/(2x - 3)] / 2 + 3
= 14/[2(2x - 3)] + 3
= 7/(2x - 3) + 3
```

Step 2: Find `f⁻¹(x)`

Let `y = x/2 + 3`

Swap x and y:

`x = y/2 + 3`

Solve for y:

`x - 3 = y/2 → y = 2(x - 3) = 2x - 6`

So `f⁻¹(x) = 2x - 6`

Step 3: Set equal

```
7/(2x - 3) + 3 = 2x - 6
```

Subtract 3 from both sides:

```
7/(2x - 3) = 2x - 9
```

Multiply both sides by `(2x - 3)` (note: `2x - 3 ≠ 0 → x ≠ 3/2`)

```
7 = (2x - 9)(2x - 3)
```

Expand right side:

```
(2x - 9)(2x - 3) = 4x² - 6x - 18x + 27 = 4x² - 24x + 27
```

So:

```
4x² - 24x + 27 = 7
→ 4x² - 24x + 20 = 0
```

Divide by 4:

```
x² - 6x + 5 = 0
→ (x - 5)(x - 1) = 0
→ x = 5 or x = 1
```

Check if these make denominator zero in original function?

Original `g(x) = 14/(2x - 3)` — at x=5: 2*5-3=7≠0; at x=1: 2*1-3=-1≠0 → OK.

✔ Answer: x = 5 or x = 1

---

## 📌 Summary of Solutions:

| Problem | Answer |
|---------|--------|
| A1 `gf(x)=21` | x = 4 or x = -1 |
| B1 `gf(x)=g(x)` | No real solutions (complex: x = (-3 ± i√3)/3) |
| C1 `fg(x)=g⁻¹(x)` | x = 6 or x = 1 |
| D1 `fg(x)=f⁻¹(x)` | x = 5 or x = 1 |

---

## 🔍 General Strategy for All Problems:

1. Understand notation: `fg(x)` means `f(g(x))`, `gf(x)` means `g(f(x))`.
2. Compute composite function by substituting one function into another.
3. If inverse is needed, solve `y = f(x)` for x, then swap variables.
4. Set up equation and solve algebraically (expand, simplify, factor, use quadratic formula).
5. Check domain restrictions (especially with rational functions or square roots).

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Let me know if you’d like me to solve another specific cell (like A2, B3, etc.) — just tell me which one!

Happy solving! 😊