Solved ID: 1 Nam College Algebra Inverse Function Worksheet ... - Free Printable

To find 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 solve each problem step by step.

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1. \( g(x) = -\frac{1}{x-1} + 3 \)

#### Step 1: Replace \( g(x) \) with \( y \):
\[ y = -\frac{1}{x-1} + 3 \]

#### Step 2: Swap \( x \) and \( y \):
\[ x = -\frac{1}{y-1} + 3 \]

#### Step 3: Solve for \( y \):
1. Isolate the fraction:
\[ x - 3 = -\frac{1}{y-1} \]

2. Take the reciprocal (and flip the sign):
\[ \frac{1}{x-3} = -(y-1) \]

3. Simplify:
\[ \frac{1}{x-3} = -y + 1 \]

4. Solve for \( y \):
\[ -y = \frac{1}{x-3} - 1 \]
\[ y = -\left( \frac{1}{x-3} - 1 \right) \]
\[ y = -\frac{1}{x-3} + 1 \]

#### Step 4: Replace \( y \) with \( g^{-1}(x) \):
\[ g^{-1}(x) = -\frac{1}{x-3} + 1 \]

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

#### Step 1: Replace \( f(x) \) with \( y \):
\[ y = x - 6 \]

#### Step 2: Swap \( x \) and \( y \):
\[ x = y - 6 \]

#### Step 3: Solve for \( y \):
\[ y = x + 6 \]

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

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3. \( g(x) = \frac{2}{5}x - 2 \)

#### Step 1: Replace \( g(x) \) with \( y \):
\[ y = \frac{2}{5}x - 2 \]

#### Step 2: Swap \( x \) and \( y \):
\[ x = \frac{2}{5}y - 2 \]

#### Step 3: Solve for \( y \):
1. Add 2 to both sides:
\[ x + 2 = \frac{2}{5}y \]

2. Multiply both sides by \( \frac{5}{2} \):
\[ y = \frac{5}{2}(x + 2) \]
\[ y = \frac{5}{2}x + 5 \]

#### Step 4: Replace \( y \) with \( g^{-1}(x) \):
\[ g^{-1}(x) = \frac{5}{2}x + 5 \]

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4. \( f(v) = \sqrt[3]{v} + 2 \)

#### Step 1: Replace \( f(v) \) with \( y \):
\[ y = \sqrt[3]{v} + 2 \]

#### Step 2: Swap \( v \) and \( y \):
\[ v = \sqrt[3]{y} + 2 \]

#### Step 3: Solve for \( y \):
1. Subtract 2 from both sides:
\[ v - 2 = \sqrt[3]{y} \]

2. Cube both sides:
\[ (v - 2)^3 = y \]

#### Step 4: Replace \( y \) with \( f^{-1}(v) \):
\[ f^{-1}(v) = (v - 2)^3 \]

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5. \( g(x) = \frac{-4 + \sqrt[3]{4x}}{2} \)

#### Step 1: Replace \( g(x) \) with \( y \):
\[ y = \frac{-4 + \sqrt[3]{4x}}{2} \]

#### Step 2: Swap \( x \) and \( y \):
\[ x = \frac{-4 + \sqrt[3]{4y}}{2} \]

#### Step 3: Solve for \( y \):
1. Multiply both sides by 2:
\[ 2x = -4 + \sqrt[3]{4y} \]

2. Add 4 to both sides:
\[ 2x + 4 = \sqrt[3]{4y} \]

3. Cube both sides:
\[ (2x + 4)^3 = 4y \]

4. Divide both sides by 4:
\[ y = \frac{(2x + 4)^3}{4} \]

#### Step 4: Replace \( y \) with \( g^{-1}(x) \):
\[ g^{-1}(x) = \frac{(2x + 4)^3}{4} \]

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6. \( f(x) = \frac{4}{5}x - 4 \)

#### Step 1: Replace \( f(x) \) with \( y \):
\[ y = \frac{4}{5}x - 4 \]

#### Step 2: Swap \( x \) and \( y \):
\[ x = \frac{4}{5}y - 4 \]

#### Step 3: Solve for \( y \):
1. Add 4 to both sides:
\[ x + 4 = \frac{4}{5}y \]

2. Multiply both sides by \( \frac{5}{4} \):
\[ y = \frac{5}{4}(x + 4) \]
\[ y = \frac{5}{4}x + 5 \]

#### Step 4: Replace \( y \) with \( f^{-1}(x) \):
\[ f^{-1}(x) = \frac{5}{4}x + 5 \]

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7. \( h(n) = \frac{1}{n-2} - 2 \)

#### Step 1: Replace \( h(n) \) with \( y \):
\[ y = \frac{1}{n-2} - 2 \]

#### Step 2: Swap \( n \) and \( y \):
\[ n = \frac{1}{y-2} - 2 \]

#### Step 3: Solve for \( y \):
1. Add 2 to both sides:
\[ n + 2 = \frac{1}{y-2} \]

2. Take the reciprocal:
\[ \frac{1}{n+2} = y - 2 \]

3. Add 2 to both sides:
\[ y = \frac{1}{n+2} + 2 \]

#### Step 4: Replace \( y \) with \( h^{-1}(n) \):
\[ h^{-1}(n) = \frac{1}{n+2} + 2 \]

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8. \( g(x) = -\sqrt[3]{x} - 3 \)

#### Step 1: Replace \( g(x) \) with \( y \):
\[ y = -\sqrt[3]{x} - 3 \]

#### Step 2: Swap \( x \) and \( y \):
\[ x = -\sqrt[3]{y} - 3 \]

#### Step 3: Solve for \( y \):
1. Add 3 to both sides:
\[ x + 3 = -\sqrt[3]{y} \]

2. Multiply both sides by -1:
\[ -(x + 3) = \sqrt[3]{y} \]

3. Cube both sides:
\[ (-x - 3)^3 = y \]

#### Step 4: Replace \( y \) with \( g^{-1}(x) \):
\[ g^{-1}(x) = (-x - 3)^3 \]

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9. \( g(x) = -2x^3 - 2 \)

#### Step 1: Replace \( g(x) \) with \( y \):
\[ y = -2x^3 - 2 \]

#### Step 2: Swap \( x \) and \( y \):
\[ x = -2y^3 - 2 \]

#### Step 3: Solve for \( y \):
1. Add 2 to both sides:
\[ x + 2 = -2y^3 \]

2. Divide both sides by -2:
\[ \frac{x + 2}{-2} = y^3 \]
\[ -\frac{x + 2}{2} = y^3 \]

3. Take the cube root:
\[ y = \sqrt[3]{-\frac{x + 2}{2}} \]

#### Step 4: Replace \( y \) with \( g^{-1}(x) \):
\[ g^{-1}(x) = \sqrt[3]{-\frac{x + 2}{2}} \]

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10. \( f(y) = -\frac{1}{x} - 1 \)

#### Step 1: Replace \( f(y) \) with \( x \):
\[ x = -\frac{1}{y} - 1 \]

#### Step 2: Swap \( x \) and \( y \):
\[ y = -\frac{1}{x} - 1 \]

#### Step 3: Solve for \( x \):
1. Add 1 to both sides:
\[ y + 1 = -\frac{1}{x} \]

2. Take the reciprocal (and flip the sign):
\[ \frac{1}{y+1} = -x \]

3. Multiply both sides by -1:
\[ x = -\frac{1}{y+1} \]

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

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Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad g^{-1}(x) = -\frac{1}{x-3} + 1 \\
2. & \quad f^{-1}(x) = x + 6 \\
3. & \quad g^{-1}(x) = \frac{5}{2}x + 5 \\
4. & \quad f^{-1}(v) = (v - 2)^3 \\
5. & \quad g^{-1}(x) = \frac{(2x + 4)^3}{4} \\
6. & \quad f^{-1}(x) = \frac{5}{4}x + 5 \\
7. & \quad h^{-1}(n) = \frac{1}{n+2} + 2 \\
8. & \quad g^{-1}(x) = (-x - 3)^3 \\
9. & \quad g^{-1}(x) = \sqrt[3]{-\frac{x + 2}{2}} \\
10. & \quad f^{-1}(y) = -\frac{1}{y+1}
\end{aligned}
}
\]