Inverse Function Worksheets - Free Printable

To solve the problem of finding the inverse of each function, we need to follow a systematic approach. The general steps are:

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 go through each function step by step.

---

Problem I: \( f(x) = 6x - 9 \)

1. Replace \( f(x) \) with \( y \):
\[
y = 6x - 9
\]

2. Swap \( x \) and \( y \):
\[
x = 6y - 9
\]

3. Solve for \( y \):
\[
x + 9 = 6y
\]
\[
y = \frac{x + 9}{6}
\]

4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \frac{x + 9}{6}
\]

Answer for I:
\[
\boxed{f^{-1}(x) = \frac{x + 9}{6}}
\]

---

Problem II: \( g(x) = (x + 5)^3 \)

1. Replace \( g(x) \) with \( y \):
\[
y = (x + 5)^3
\]

2. Swap \( x \) and \( y \):
\[
x = (y + 5)^3
\]

3. Solve for \( y \):
\[
\sqrt[3]{x} = y + 5
\]
\[
y = \sqrt[3]{x} - 5
\]

4. Replace \( y \) with \( g^{-1}(x) \):
\[
g^{-1}(x) = \sqrt[3]{x} - 5
\]

Answer for II:
\[
\boxed{g^{-1}(x) = \sqrt[3]{x} - 5}
\]

---

Problem III: \( h(x) = \frac{x}{7} \)

1. Replace \( h(x) \) with \( y \):
\[
y = \frac{x}{7}
\]

2. Swap \( x \) and \( y \):
\[
x = \frac{y}{7}
\]

3. Solve for \( y \):
\[
y = 7x
\]

4. Replace \( y \) with \( h^{-1}(x) \):
\[
h^{-1}(x) = 7x
\]

Answer for III:
\[
\boxed{h^{-1}(x) = 7x}
\]

---

Problem IV: \( j(x) = 30 - x \)

1. Replace \( j(x) \) with \( y \):
\[
y = 30 - x
\]

2. Swap \( x \) and \( y \):
\[
x = 30 - y
\]

3. Solve for \( y \):
\[
y = 30 - x
\]

4. Replace \( y \) with \( j^{-1}(x) \):
\[
j^{-1}(x) = 30 - x
\]

Answer for IV:
\[
\boxed{j^{-1}(x) = 30 - x}
\]

---

Problem V: \( k(x) = mx + t \)

1. Replace \( k(x) \) with \( y \):
\[
y = mx + t
\]

2. Swap \( x \) and \( y \):
\[
x = my + t
\]

3. Solve for \( y \):
\[
x - t = my
\]
\[
y = \frac{x - t}{m}
\]

4. Replace \( y \) with \( k^{-1}(x) \):
\[
k^{-1}(x) = \frac{x - t}{m}
\]

Answer for V:
\[
\boxed{k^{-1}(x) = \frac{x - t}{m}}
\]

---

Problem VI: \( l(x) = \frac{5}{2} + \frac{x}{5} \)

1. Replace \( l(x) \) with \( y \):
\[
y = \frac{5}{2} + \frac{x}{5}
\]

2. Swap \( x \) and \( y \):
\[
x = \frac{5}{2} + \frac{y}{5}
\]

3. Solve for \( y \):
\[
x - \frac{5}{2} = \frac{y}{5}
\]
\[
y = 5 \left( x - \frac{5}{2} \right)
\]
\[
y = 5x - \frac{25}{2}
\]

4. Replace \( y \) with \( l^{-1}(x) \):
\[
l^{-1}(x) = 5x - \frac{25}{2}
\]

Answer for VI:
\[
\boxed{l^{-1}(x) = 5x - \frac{25}{2}}
\]

---

Problem VII: \( m(x) = \frac{3}{4}x \)

1. Replace \( m(x) \) with \( y \):
\[
y = \frac{3}{4}x
\]

2. Swap \( x \) and \( y \):
\[
x = \frac{3}{4}y
\]

3. Solve for \( y \):
\[
y = \frac{4}{3}x
\]

4. Replace \( y \) with \( m^{-1}(x) \):
\[
m^{-1}(x) = \frac{4}{3}x
\]

Answer for VII:
\[
\boxed{m^{-1}(x) = \frac{4}{3}x}
\]

---

Problem VIII: \( n(x) = 7x - 38 \)

1. Replace \( n(x) \) with \( y \):
\[
y = 7x - 38
\]

2. Swap \( x \) and \( y \):
\[
x = 7y - 38
\]

3. Solve for \( y \):
\[
x + 38 = 7y
\]
\[
y = \frac{x + 38}{7}
\]

4. Replace \( y \) with \( n^{-1}(x) \):
\[
n^{-1}(x) = \frac{x + 38}{7}
\]

Answer for VIII:
\[
\boxed{n^{-1}(x) = \frac{x + 38}{7}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{I: } f^{-1}(x) = \frac{x + 9}{6} \\
&\text{II: } g^{-1}(x) = \sqrt[3]{x} - 5 \\
&\text{III: } h^{-1}(x) = 7x \\
&\text{IV: } j^{-1}(x) = 30 - x \\
&\text{V: } k^{-1}(x) = \frac{x - t}{m} \\
&\text{VI: } l^{-1}(x) = 5x - \frac{25}{2} \\
&\text{VII: } m^{-1}(x) = \frac{4}{3}x \\
&\text{VIII: } n^{-1}(x) = \frac{x + 38}{7}
\end{aligned}
}
\]