Inverse Functions (A) Worksheet | Fun and Engaging Algebra II PDF ... - Free Printable

Let's solve each problem step-by-step.

---

Section A: Find the inverse of each function

To find the inverse of a function $ f(x) $, 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) $.

---

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

- $ y = x + 4 $
- Swap: $ x = y + 4 $
- Solve: $ y = x - 4 $

✔ Answer: $ f^{-1}(x) = x - 4 $

---

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

- $ y = 6x - 2 $
- Swap: $ x = 6y - 2 $
- Solve: $ x + 2 = 6y $ → $ y = \frac{x + 2}{6} $

✔ Answer: $ f^{-1}(x) = \frac{x + 2}{6} $

---

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

- $ y = \frac{x}{8} $
- Swap: $ x = \frac{y}{8} $
- Solve: $ y = 8x $

✔ Answer: $ f^{-1}(x) = 8x $

---

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

- $ y = \frac{x}{2} - 7 $
- Swap: $ x = \frac{y}{2} - 7 $
- Solve: $ x + 7 = \frac{y}{2} $ → $ y = 2(x + 7) = 2x + 14 $

✔ Answer: $ f^{-1}(x) = 2x + 14 $

---

#### 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 $
- Solve: $ 4x + 37 = -5y $ → $ y = \frac{-4x - 37}{5} $

✔ Answer: $ f^{-1}(x) = \frac{-4x - 37}{5} $

---

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

Note: This is not one-to-one over all real numbers (fails horizontal line test), so we must restrict domain to make it invertible.

Assume $ x \geq 0 $ (or $ x \leq 0 $). We'll assume $ x \geq 0 $.

- $ y = x^2 - 10 $
- Swap: $ x = y^2 - 10 $
- Solve: $ y^2 = x + 10 $ → $ y = \sqrt{x + 10} $ (since $ y \geq 0 $)

✔ Answer: $ f^{-1}(x) = \sqrt{x + 10} $, with domain $ x \geq -10 $

---

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

This is also quadratic in nature, so not one-to-one unless domain is restricted.

Assume $ x \geq 0 $ (for simplicity).

- $ y = \frac{2x^2 + 9}{15} $
- Swap: $ x = \frac{2y^2 + 9}{15} $
- Multiply: $ 15x = 2y^2 + 9 $
- $ 2y^2 = 15x - 9 $
- $ y^2 = \frac{15x - 9}{2} $
- $ y = \sqrt{\frac{15x - 9}{2}} $ (positive root since $ y \geq 0 $)

✔ Answer: $ f^{-1}(x) = \sqrt{\frac{15x - 9}{2}} $, domain $ x \geq \frac{9}{15} = 0.6 $

---

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

Domain: $ 4x + 13 \geq 0 $ → $ x \geq -\frac{13}{4} $

- $ y = \sqrt{4x + 13} $
- Swap: $ x = \sqrt{4y + 13} $
- Square both sides: $ x^2 = 4y + 13 $
- $ 4y = x^2 - 13 $
- $ y = \frac{x^2 - 13}{4} $

But note: original function outputs $ y \geq 0 $, so inverse domain is $ x \geq 0 $

✔ Answer: $ f^{-1}(x) = \frac{x^2 - 13}{4} $, domain $ x \geq 0 $

---

Section B

---

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

We are solving $ f^{-1}(x) = 3 $

This means: $ f(3) = x $

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

✔ Answer: $ x = 38 $

---

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

We need to complete the square or use algebra.

- $ y = x^2 + 4x + 3 $
- Swap: $ x = y^2 + 4y + 3 $
- But this is messy — better to solve for $ y $ in terms of $ x $.

Start again:

- $ y = x^2 + 4x + 3 $
- Complete the square:
$$
y = (x^2 + 4x + 4) - 1 = (x + 2)^2 - 1
$$

So:
$$
y = (x + 2)^2 - 1
$$

Now swap:
$$
x = (y + 2)^2 - 1
$$
$$
x + 1 = (y + 2)^2
$$
$$
y + 2 = \pm \sqrt{x + 1}
$$
$$
y = -2 \pm \sqrt{x + 1}
$$

But since $ x > 0 $, we need to determine which sign to take.

Original function: $ f(x) = x^2 + 4x + 3 $, increasing for $ x > 0 $, and $ f(0) = 3 $, so range starts at 3.

As $ x \to \infty $, $ f(x) \to \infty $. So we want positive branch.

But $ y = -2 + \sqrt{x + 1} $ or $ y = -2 - \sqrt{x + 1} $

Try $ x = 1 $: $ f(1) = 1 + 4 + 3 = 8 $

Then $ f^{-1}(8) = ? $

Try $ y = -2 + \sqrt{8 + 1} = -2 + 3 = 1 $ ✔

Other branch: $ -2 - 3 = -5 $ ✘

So we take positive root.

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

---

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

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

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

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

Solve for $ y $:

Multiply both sides:
$$
x(y - 2) = 2 + 3y
$$
$$
xy - 2x = 2 + 3y
$$
Bring all terms to one side:
$$
xy - 3y = 2 + 2x
$$
Factor:
$$
y(x - 3) = 2 + 2x
$$
$$
y = \frac{2x + 2}{x - 3}
$$

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

---

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

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

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

✔ Answer: $ f^{-1}(g(-2)) = 10 $

---

Extension

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

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

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

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

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

Multiply: $ 3x = 2y - 5 $

$ 2y = 3x + 5 $

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

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

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

Multiply both sides by $ 2(2 - x) $ to eliminate denominators:

$$
(3x + 5)(2 - x) = 2 \cdot 4 = 8
$$

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

Set equal:
$$
-3x^2 + x + 10 = 8
\Rightarrow -3x^2 + x + 2 = 0
$$

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 when $ x = 2 $, so both values OK.

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

---

✔ Final Answers Summary

---

#### 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}, x \geq -10 $ |
| 7) $ f(x) = \frac{2x^2 + 9}{15} $ | $ f^{-1}(x) = \sqrt{\frac{15x - 9}{2}}, x \geq 0.6 $ |
| 8) $ f(x) = \sqrt{4x + 13} $ | $ f^{-1}(x) = \frac{x^2 - 13}{4}, x \geq 0 $ |

---

#### Section B

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

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

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

---

#### Extension

Solve $ f^{-1}(x) = g(x) $:
Solutions: $ x = -\frac{2}{3} $ and $ x = 1 $

---

Let me know if you'd like this formatted as a downloadable PDF or printed worksheet!