Inverse Functions (A) Worksheet | Cazoom Maths Worksheets - Free Printable

Let's solve the Inverse Functions (A) worksheet step by step, explaining each solution clearly.

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🔷 Section A: Find the inverse of each function

To find the inverse function $ f^{-1}(x) $, we follow these steps:
1. Replace $ f(x) $ with $ y $.
2. Swap $ x $ and $ y $.
3. Solve for $ y $.
4. Replace $ y $ with $ f^{-1}(x) $.

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#### 1) $ f(x) = x + 4 $

- $ y = x + 4 $
- Swap: $ x = y + 4 $
- Solve: $ y = x - 4 $
✔ $ f^{-1}(x) = x - 4 $

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#### 2) $ f(x) = 6x - 2 $

- $ y = 6x - 2 $
- Swap: $ x = 6y - 2 $
- Add 2: $ x + 2 = 6y $
- Divide: $ y = \frac{x + 2}{6} $
✔ $ f^{-1}(x) = \frac{x + 2}{6} $

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#### 3) $ f(x) = \frac{x}{8} $

- $ y = \frac{x}{8} $
- Swap: $ x = \frac{y}{8} $
- Multiply: $ y = 8x $
✔ $ f^{-1}(x) = 8x $

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#### 4) $ f(x) = \frac{x}{2} - 7 $

- $ y = \frac{x}{2} - 7 $
- Swap: $ x = \frac{y}{2} - 7 $
- Add 7: $ x + 7 = \frac{y}{2} $
- Multiply: $ y = 2(x + 7) = 2x + 14 $
✔ $ f^{-1}(x) = 2x + 14 $

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#### 5) $ f(x) = \frac{11 - 5x}{4} - 12 $

First simplify:
$$
f(x) = \frac{11 - 5x}{4} - 12 = \frac{11 - 5x - 48}{4} = \frac{-5x - 37}{4}
$$

Now find inverse:

- $ y = \frac{-5x - 37}{4} $
- Swap: $ x = \frac{-5y - 37}{4} $
- Multiply both sides by 4: $ 4x = -5y - 37 $
- Add 37: $ 4x + 37 = -5y $
- Divide: $ y = \frac{-(4x + 37)}{5} = -\frac{4x + 37}{5} $
✔ $ f^{-1}(x) = -\frac{4x + 37}{5} $

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#### 6) $ f(x) = x^2 - 10 $

This is a quadratic, so it's not one-to-one unless domain is restricted.

But since no restriction is given, we assume $ x \geq 0 $ (common convention for square roots).

- $ y = x^2 - 10 $
- Swap: $ x = y^2 - 10 $
- Add 10: $ x + 10 = y^2 $
- Take square root: $ y = \sqrt{x + 10} $ (since $ y \geq 0 $)
✔ $ f^{-1}(x) = \sqrt{x + 10} $

> Note: If $ x < 0 $ were allowed, there would be two inverses, but typically we take the principal (positive) root.

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#### 7) $ f(x) = \frac{2x^2 + 9}{15} $

- $ y = \frac{2x^2 + 9}{15} $
- Swap: $ x = \frac{2y^2 + 9}{15} $
- Multiply: $ 15x = 2y^2 + 9 $
- Subtract 9: $ 15x - 9 = 2y^2 $
- Divide: $ y^2 = \frac{15x - 9}{2} $
- Take square root: $ y = \pm \sqrt{\frac{15x - 9}{2}} $

But again, to have an inverse function, we must restrict domain.

Assuming $ x \geq 0 $, then $ y \geq 0 $, so:

✔ $ f^{-1}(x) = \sqrt{\frac{15x - 9}{2}} $ (principal root)

> Note: This is only defined when $ 15x - 9 \geq 0 \Rightarrow x \geq \frac{3}{5} $

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#### 8) $ f(x) = \sqrt{4x + 13} $

- $ y = \sqrt{4x + 13} $
- Swap: $ x = \sqrt{4y + 13} $
- Square both sides: $ x^2 = 4y + 13 $
- Subtract 13: $ x^2 - 13 = 4y $
- Divide: $ y = \frac{x^2 - 13}{4} $
✔ $ f^{-1}(x) = \frac{x^2 - 13}{4} $

> Domain: $ x \geq 0 $ because $ \sqrt{...} \geq 0 $

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✔ Section A Summary:

| Function | Inverse |
|--------|--------|
| 1) $ f(x) = x + 4 $ | $ f^{-1}(x) = x - 4 $ |
| 2) $ f(x) = 6x - 2 $ | $ f^{-1}(x) = \frac{x + 2}{6} $ |
| 3) $ f(x) = \frac{x}{8} $ | $ f^{-1}(x) = 8x $ |
| 4) $ f(x) = \frac{x}{2} - 7 $ | $ f^{-1}(x) = 2x + 14 $ |
| 5) $ f(x) = \frac{11 - 5x}{4} - 12 $ | $ f^{-1}(x) = -\frac{4x + 37}{5} $ |
| 6) $ f(x) = x^2 - 10 $ | $ f^{-1}(x) = \sqrt{x + 10} $ |
| 7) $ f(x) = \frac{2x^2 + 9}{15} $ | $ f^{-1}(x) = \sqrt{\frac{15x - 9}{2}} $ |
| 8) $ f(x) = \sqrt{4x + 13} $ | $ f^{-1}(x) = \frac{x^2 - 13}{4} $ |

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🔷 Section B

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#### 1) Let $ f(x) = 2x^3 - 16 $. Solve $ f^{-1}(x) = 3 $

We are told $ f^{-1}(x) = 3 $. That means:

$$
f^{-1}(x) = 3 \Rightarrow f(3) = x
$$

So compute $ f(3) $:

$$
f(3) = 2(3)^3 - 16 = 2(27) - 16 = 54 - 16 = 38
$$

✔ So $ x = 38 $

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#### 2) $ f(x) = x^2 + 4x + 3 $, $ x > 0 $. Find $ f^{-1}(x) $

Complete the square:

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

Now find inverse:

- $ y = (x+2)^2 - 1 $
- Swap: $ x = (y+2)^2 - 1 $
- Add 1: $ x + 1 = (y+2)^2 $
- Take square root: $ y + 2 = \sqrt{x + 1} $ (since $ x > 0 $, $ y > -2 $? But need care)

Wait: Since $ x > 0 $, let’s check values:

At $ x = 0 $, $ f(0) = 3 $

As $ x $ increases, $ f(x) $ increases.

Minimum at $ x = -2 $, but since $ x > 0 $, $ f(x) $ is increasing from $ f(0) = 3 $ onward.

So $ f(x) > 3 $ for $ x > 0 $

Now solve:

- $ x = (y+2)^2 - 1 $
- $ x + 1 = (y+2)^2 $
- $ y + 2 = \sqrt{x + 1} $ → since $ y > -2 $, we take positive root
- $ y = \sqrt{x + 1} - 2 $

✔ $ f^{-1}(x) = \sqrt{x + 1} - 2 $, for $ x > 3 $

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#### 3) Let $ f(x) = \frac{2 + 3x}{x - 2} $, $ g(x) = x^2 $

##### a) Find the inverse of $ f(x) $

Set $ y = \frac{2 + 3x}{x - 2} $

Swap $ x $ and $ y $:

$$
x = \frac{2 + 3y}{y - 2}
$$

Solve for $ y $:

Multiply both sides by $ y - 2 $:

$$
x(y - 2) = 2 + 3y \\
xy - 2x = 2 + 3y \\
xy - 3y = 2 + 2x \\
y(x - 3) = 2 + 2x \\
y = \frac{2 + 2x}{x - 3}
$$

✔ $ f^{-1}(x) = \frac{2 + 2x}{x - 3} $

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##### b) Find $ f^{-1}(g(-2)) $

First, $ g(-2) = (-2)^2 = 4 $

Then $ f^{-1}(4) = \frac{2 + 2(4)}{4 - 3} = \frac{2 + 8}{1} = 10 $

✔ $ f^{-1}(g(-2)) = 10 $

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🔷 Extension

Given:
- $ f(x) = \frac{1}{3}(2x - 5) $
- $ g(x) = \frac{4}{2 - x} $

Solve $ f^{-1}(x) = g(x) $

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Step 1: Find $ f^{-1}(x) $

$ f(x) = \frac{1}{3}(2x - 5) $

Let $ y = \frac{1}{3}(2x - 5) $

Swap: $ x = \frac{1}{3}(2y - 5) $

Multiply by 3: $ 3x = 2y - 5 $

Add 5: $ 3x + 5 = 2y $

Divide: $ y = \frac{3x + 5}{2} $

So $ f^{-1}(x) = \frac{3x + 5}{2} $

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Now solve:

$$
f^{-1}(x) = g(x) \Rightarrow \frac{3x + 5}{2} = \frac{4}{2 - x}
$$

Cross-multiply:

$$
(3x + 5)(2 - x) = 8
$$

Expand left:

$$
3x(2 - x) + 5(2 - x) = 6x - 3x^2 + 10 - 5x = -3x^2 + x + 10
$$

Set equal:

$$
-3x^2 + x + 10 = 8 \\
-3x^2 + x + 2 = 0 \\
\text{Multiply by } -1: 3x^2 - x - 2 = 0
$$

Factor:

$$
(3x + 2)(x - 1) = 0
$$

So $ x = -\frac{2}{3} $ or $ x = 1 $

Check for restrictions:

- $ g(x) = \frac{4}{2 - x} $ → undefined at $ x = 2 $
- Both solutions are valid

✔ Solutions: $ x = -\frac{2}{3} $ and $ x = 1 $

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✔ Final Answers

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Section A

| Function | Inverse |
|--------|--------|
| 1) $ f(x) = x + 4 $ | $ f^{-1}(x) = x - 4 $ |
| 2) $ f(x) = 6x - 2 $ | $ f^{-1}(x) = \frac{x + 2}{6} $ |
| 3) $ f(x) = \frac{x}{8} $ | $ f^{-1}(x) = 8x $ |
| 4) $ f(x) = \frac{x}{2} - 7 $ | $ f^{-1}(x) = 2x + 14 $ |
| 5) $ f(x) = \frac{11 - 5x}{4} - 12 $ | $ f^{-1}(x) = -\frac{4x + 37}{5} $ |
| 6) $ f(x) = x^2 - 10 $ | $ f^{-1}(x) = \sqrt{x + 10} $ |
| 7) $ f(x) = \frac{2x^2 + 9}{15} $ | $ f^{-1}(x) = \sqrt{\frac{15x - 9}{2}} $ |
| 8) $ f(x) = \sqrt{4x + 13} $ | $ f^{-1}(x) = \frac{x^2 - 13}{4} $ |

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Section B

1) $ f^{-1}(x) = 3 \Rightarrow x = \boxed{38} $

2) $ f^{-1}(x) = \boxed{\sqrt{x + 1} - 2} $, for $ x > 3 $

3)
a) $ f^{-1}(x) = \boxed{\frac{2 + 2x}{x - 3}} $
b) $ f^{-1}(g(-2)) = \boxed{10} $

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Extension

Solve $ f^{-1}(x) = g(x) $

Answer: $ \boxed{x = -\frac{2}{3}} $ or $ \boxed{x = 1} $

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