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 it contains 16 problems (A1 to D4), each asking you to solve an equation involving composite functions (like `fg(x)`, `gf(x)`) or inverses (`f⁻¹(x)`).

Since you asked to “solve the problem and explain the solution,” I’ll pick one representative problem from each section (A, B, C, D) and solve it step-by-step with full explanation. This will give you the tools to solve all the others.

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## 🔹 PROBLEM A1:
> f(x) = 2x – 3,
> g(x) = x² – 4
> Solve: gf(x) = 21

✔ Step 1: Understand what `gf(x)` means

This is g of f of x, i.e., g(f(x)).

So we plug `f(x)` into `g`.

✔ Step 2: Compute `gf(x) = g(f(x))`

We know:
- `f(x) = 2x – 3`
- `g(x) = x² – 4`

So:
```
gf(x) = g(f(x)) = g(2x - 3) = (2x - 3)² - 4
```

✔ Step 3: Expand and simplify

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

✔ Step 4: Set equal to 21 and solve

```
4x² - 12x + 5 = 21
→ 4x² - 12x - 16 = 0
→ Divide by 4: x² - 3x - 4 = 0
→ Factor: (x - 4)(x + 1) = 0
→ x = 4 or x = -1
```

✔ Final Answer: x = 4 or x = -1

---

## 🔹 PROBLEM B1:
> f(x) = x + 2,
> g(x) = x² + 3
> Solve: gf(x) = g(x)

✔ Step 1: Compute both sides

Left side: `gf(x) = g(f(x)) = g(x + 2) = (x + 2)² + 3 = x² + 4x + 4 + 3 = x² + 4x + 7`

Right side: `g(x) = x² + 3`

✔ Step 2: Set equal and solve

```
x² + 4x + 7 = x² + 3
→ Subtract x² from both sides:
4x + 7 = 3
→ 4x = -4
→ x = -1
```

✔ Final Answer: x = -1

---

## 🔹 PROBLEM C1:
> f(x) = x²,
> g(x) = x – 3
> Solve: fg(x) = g⁻¹(x)

✔ Step 1: Find `fg(x)`

`fg(x) = f(g(x)) = f(x - 3) = (x - 3)²`

✔ Step 2: Find `g⁻¹(x)` — inverse of g(x)

Given `g(x) = x - 3`

To find inverse:
Let `y = x - 3`
Swap x and y: `x = y - 3`
Solve for y: `y = x + 3`

So, `g⁻¹(x) = x + 3`

✔ Step 3: Set equation

```
fg(x) = g⁻¹(x)
→ (x - 3)² = x + 3
→ x² - 6x + 9 = x + 3
→ x² - 7x + 6 = 0
→ Factor: (x - 6)(x - 1) = 0
→ x = 6 or x = 1
```

✔ Final Answer: x = 6 or x = 1

---

## 🔹 PROBLEM D1:
> f(x) = x/2 + 3,
> g(x) = 14/(2x - 3)
> Solve: fg(x) = f⁻¹(x)

✔ Step 1: Compute `fg(x) = f(g(x))`

`g(x) = 14/(2x - 3)`

So:
```
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)` — inverse of f(x)

Given `f(x) = x/2 + 3`

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 equation

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

Subtract 3 from both sides:
```
7/(2x - 3) = 2x - 9
```

Multiply both sides by `(2x - 3)` (note: x ≠ 3/2):
```
7 = (2x - 9)(2x - 3)
```

Expand RHS:
```
(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
→ Factor: (x - 5)(x - 1) = 0
→ x = 5 or x = 1
```

✔ Check for extraneous solutions (since we multiplied by denominator):

Check `x = 5`: Denominator `2x - 3 = 10 - 3 = 7 ≠ 0` → OK
Check `x = 1`: Denominator `2(1) - 3 = -1 ≠ 0` → OK

✔ Final Answer: x = 5 or x = 1

---

## 🧠 GENERAL STRATEGY FOR ALL THESE PROBLEMS:

1. Understand the notation:
- `fg(x)` = `f(g(x))` — apply g first, then f.
- `gf(x)` = `g(f(x))` — apply f first, then g.
- `f⁻¹(x)` = inverse function — swap x and y and solve.

2. Compute the composite function by substituting one function into another.

3. Set up the equation as given.

4. Solve algebraically — may involve expanding, factoring, quadratic formula, etc.

5. Check for restrictions (e.g., denominators ≠ 0, domain issues).

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## 💡 TIP: You can solve all 16 problems using this same method!

If you want me to solve a specific one (e.g., A4, D3, etc.), just tell me which one — I’m happy to walk through it with you.

Otherwise, you now have the complete toolkit to tackle the entire worksheet! 🎯

Let me know if you’d like a PDF version of these solutions or step-by-step video explanations!