Inverse Functions worksheet for GCSE Higher Tier, featuring problems on finding inverses of various algebraic functions.

Section A: Find the inverse of each function

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

Let's solve each function step by step.

---

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

1. Replace \( f(x) \) with \( y \):
\[
y = x + 4
\]
2. Swap \( x \) and \( y \):
\[
x = y + 4
\]
3. Solve for \( y \):
\[
y = x - 4
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = x - 4
\]

Solution: \( f^{-1}(x) = x - 4 \)

---

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

1. Replace \( f(x) \) with \( y \):
\[
y = 6x - 2
\]
2. Swap \( x \) and \( y \):
\[
x = 6y - 2
\]
3. Solve for \( y \):
\[
x + 2 = 6y \implies y = \frac{x + 2}{6}
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \frac{x + 2}{6}
\]

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

---

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

1. Replace \( f(x) \) with \( y \):
\[
y = \frac{x}{8}
\]
2. Swap \( x \) and \( y \):
\[
x = \frac{y}{8}
\]
3. Solve for \( y \):
\[
y = 8x
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = 8x
\]

Solution: \( f^{-1}(x) = 8x \)

---

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

1. Replace \( f(x) \) with \( y \):
\[
y = \frac{x}{2} - 7
\]
2. Swap \( x \) and \( y \):
\[
x = \frac{y}{2} - 7
\]
3. Solve for \( y \):
\[
x + 7 = \frac{y}{2} \implies y = 2(x + 7) = 2x + 14
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = 2x + 14
\]

Solution: \( f^{-1}(x) = 2x + 14 \)

---

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

1. Replace \( f(x) \) with \( y \):
\[
y = \frac{11 - 5x}{4} - 12
\]
2. Simplify the expression:
\[
y = \frac{11 - 5x}{4} - \frac{48}{4} = \frac{11 - 5x - 48}{4} = \frac{-5x - 37}{4}
\]
3. Swap \( x \) and \( y \):
\[
x = \frac{-5y - 37}{4}
\]
4. Solve for \( y \):
\[
4x = -5y - 37 \implies -5y = 4x + 37 \implies y = -\frac{4x + 37}{5}
\]
5. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = -\frac{4x + 37}{5}
\]

Solution: \( f^{-1}(x) = -\frac{4x + 37}{5} \)

---

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

This function is not one-to-one because it is a parabola opening upwards. To find an inverse, we need to restrict the domain. Assuming \( x \geq 0 \) (the principal square root):

1. Replace \( f(x) \) with \( y \):
\[
y = x^2 - 10
\]
2. Swap \( x \) and \( y \):
\[
x = y^2 - 10
\]
3. Solve for \( y \):
\[
y^2 = x + 10 \implies y = \sqrt{x + 10} \quad (\text{since } y \geq 0)
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \sqrt{x + 10}
\]

Solution: \( f^{-1}(x) = \sqrt{x + 10} \)

---

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

This function is not one-to-one because it is a parabola opening upwards. To find an inverse, we need to restrict the domain. Assuming \( x \geq 0 \) (the principal square root):

1. Replace \( f(x) \) with \( y \):
\[
y = \frac{2x^2 + 9}{15}
\]
2. Swap \( x \) and \( y \):
\[
x = \frac{2y^2 + 9}{15}
\]
3. Solve for \( y \):
\[
15x = 2y^2 + 9 \implies 2y^2 = 15x - 9 \implies y^2 = \frac{15x - 9}{2} \implies y = \sqrt{\frac{15x - 9}{2}} \quad (\text{since } y \geq 0)
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \sqrt{\frac{15x - 9}{2}}
\]

Solution: \( f^{-1}(x) = \sqrt{\frac{15x - 9}{2}} \)

---

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

1. Replace \( f(x) \) with \( y \):
\[
y = \sqrt{4x + 13}
\]
2. Swap \( x \) and \( y \):
\[
x = \sqrt{4y + 13}
\]
3. Solve for \( y \):
\[
x^2 = 4y + 13 \implies 4y = x^2 - 13 \implies y = \frac{x^2 - 13}{4}
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \frac{x^2 - 13}{4}
\]

Solution: \( f^{-1}(x) = \frac{x^2 - 13}{4} \)

---

Section B

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

To solve \( f^{-1}(x) = 3 \), we need to find \( x \) such that \( f(3) = x \).

1. Substitute \( x = 3 \) into \( f(x) \):
\[
f(3) = 2(3)^3 - 16 = 2(27) - 16 = 54 - 16 = 38
\]

Solution: \( x = 38 \)

---

#### 2) When \( f(x) = x^2 + 4x + 3 \), \( x > 0 \), find \( f^{-1}(x) \).

First, complete the square for \( f(x) \):

1. Rewrite \( f(x) \):
\[
f(x) = x^2 + 4x + 3 = (x + 2)^2 - 4 + 3 = (x + 2)^2 - 1
\]
2. Since \( x > 0 \), we have \( x + 2 > 2 \). Thus, \( f(x) \) is one-to-one on this domain.
3. Replace \( f(x) \) with \( y \):
\[
y = (x + 2)^2 - 1
\]
4. Swap \( x \) and \( y \):
\[
x = (y + 2)^2 - 1
\]
5. Solve for \( y \):
\[
x + 1 = (y + 2)^2 \implies y + 2 = \sqrt{x + 1} \quad (\text{since } y + 2 > 0) \implies y = \sqrt{x + 1} - 2
\]
6. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \sqrt{x + 1} - 2
\]

Solution: \( f^{-1}(x) = \sqrt{x + 1} - 2 \)

---

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

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

1. Replace \( f(x) \) with \( y \):
\[
y = \frac{2 + 3x}{x - 2}
\]
2. Swap \( x \) and \( y \):
\[
x = \frac{2 + 3y}{y - 2}
\]
3. Solve for \( y \):
\[
x(y - 2) = 2 + 3y \implies xy - 2x = 2 + 3y \implies xy - 3y = 2 + 2x \implies y(x - 3) = 2 + 2x \implies y = \frac{2 + 2x}{x - 3}
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \frac{2 + 2x}{x - 3}
\]

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

##### b) Find the value of \( f^{-1}(g(-2)) \)

1. Compute \( g(-2) \):
\[
g(-2) = (-2)^2 = 4
\]
2. Compute \( f^{-1}(4) \):
\[
f^{-1}(4) = \frac{2 + 2(4)}{4 - 3} = \frac{2 + 8}{1} = 10
\]

Solution: \( f^{-1}(g(-2)) = 10 \)

---

Extension

#### Solve the equation \( f^{-1}(x) = g(x) \)

Given:
\[
f(x) = \frac{1}{3}(2x - 5) \quad \text{and} \quad g(x) = \frac{4}{2 - x}
\]

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

1. Replace \( f(x) \) with \( y \):
\[
y = \frac{1}{3}(2x - 5)
\]
2. Swap \( x \) and \( y \):
\[
x = \frac{1}{3}(2y - 5)
\]
3. Solve for \( y \):
\[
3x = 2y - 5 \implies 2y = 3x + 5 \implies y = \frac{3x + 5}{2}
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \frac{3x + 5}{2}
\]

##### Step 2: Solve \( f^{-1}(x) = g(x) \)

We have:
\[
f^{-1}(x) = \frac{3x + 5}{2} \quad \text{and} \quad g(x) = \frac{4}{2 - x}
\]
Set them equal:
\[
\frac{3x + 5}{2} = \frac{4}{2 - x}
\]
Cross-multiply:
\[
(3x + 5)(2 - x) = 4 \cdot 2
\]
Expand and simplify:
\[
6x - 3x^2 + 10 - 5x = 8 \implies -3x^2 + x + 10 = 8 \implies -3x^2 + x + 2 = 0
\]
Multiply through by \(-1\):
\[
3x^2 - x - 2 = 0
\]
Factorize:
\[
(3x + 2)(x - 1) = 0
\]
Solve for \( x \):
\[
x = -\frac{2}{3} \quad \text{or} \quad x = 1
\]

##### Step 3: Verify solutions

- For \( x = -\frac{2}{3} \):
\[
f^{-1}\left(-\frac{2}{3}\right) = \frac{3\left(-\frac{2}{3}\right) + 5}{2} = \frac{-2 + 5}{2} = \frac{3}{2}
\]
\[
g\left(-\frac{2}{3}\right) = \frac{4}{2 - \left(-\frac{2}{3}\right)} = \frac{4}{2 + \frac{2}{3}} = \frac{4}{\frac{6}{3} + \frac{2}{3}} = \frac{4}{\frac{8}{3}} = \frac{4 \cdot 3}{8} = \frac{3}{2}
\]
Both sides are equal.

- For \( x = 1 \):
\[
f^{-1}(1) = \frac{3(1) + 5}{2} = \frac{3 + 5}{2} = 4
\]
\[
g(1) = \frac{4}{2 - 1} = \frac{4}{1} = 4
\]
Both sides are equal.

Both solutions are valid.

Solution: \( x = -\frac{2}{3} \) or \( x = 1 \)

---

Final Answers

\[
\boxed{
\begin{array}{ll}
\text{Section A:} & \\
1) & f^{-1}(x) = x - 4 \\
2) & f^{-1}(x) = \frac{x + 2}{6} \\
3) & f^{-1}(x) = 8x \\
4) & f^{-1}(x) = 2x + 14 \\
5) & f^{-1}(x) = -\frac{4x + 37}{5} \\
6) & f^{-1}(x) = \sqrt{x + 10} \\
7) & f^{-1}(x) = \sqrt{\frac{15x - 9}{2}} \\
8) & f^{-1}(x) = \frac{x^2 - 13}{4} \\
\\
\text{Section B:} & \\
1) & x = 38 \\
2) & f^{-1}(x) = \sqrt{x + 1} - 2 \\
3) & \text{a) } f^{-1}(x) = \frac{2 + 2x}{x - 3}, \text{ b) } f^{-1}(g(-2)) = 10 \\
\\
\text{Extension:} & x = -\frac{2}{3} \text{ or } x = 1
\end{array}
}
\]