Inverse Functions (harder) worksheet with solutions.

To find the inverse functions for each given function, we need to 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 part step by step.

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

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

Answer:
\[
\boxed{f^{-1}(x) = \frac{x - 1}{4}}
\]

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(b) \( f(x) = \frac{3x - 1}{x + 2} \)

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

Answer:
\[
\boxed{f^{-1}(x) = \frac{-2x - 1}{x - 3}}
\]

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(c) \( f(x) = \sqrt{x - 1} \)

1. Start with \( y = \sqrt{x - 1} \).
2. Swap \( x \) and \( y \): \( x = \sqrt{y - 1} \).
3. Solve for \( y \):
\[
x = \sqrt{y - 1} \implies x^2 = y - 1 \implies y = x^2 + 1
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = x^2 + 1
\]
Note: Since \( f(x) = \sqrt{x - 1} \) is defined for \( x \geq 1 \), the domain of \( f^{-1}(x) \) is \( x \geq 0 \).

Answer:
\[
\boxed{f^{-1}(x) = x^2 + 1, \, x \geq 0}
\]

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(d) \( f(x) = \frac{x + 1}{x - 2}, \, x \in (2, +\infty) \)

1. Start with \( y = \frac{x + 1}{x - 2} \).
2. Swap \( x \) and \( y \): \( x = \frac{y + 1}{y - 2} \).
3. Solve for \( y \):
\[
x(y - 2) = y + 1 \implies xy - 2x = y + 1 \implies xy - y = 2x + 1 \implies y(x - 1) = 2x + 1 \implies y = \frac{2x + 1}{x - 1}
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \frac{2x + 1}{x - 1}
\]
Note: The domain of \( f(x) \) is \( x \in (2, +\infty) \), so the range of \( f(x) \) is \( y \in (1, +\infty) \). Therefore, the domain of \( f^{-1}(x) \) is \( x \in (1, +\infty) \).

Answer:
\[
\boxed{f^{-1}(x) = \frac{2x + 1}{x - 1}, \, x \in (1, +\infty)}
\]

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(e) \( f(x) = \frac{1}{x - 5}, \, x \in (5, +\infty) \)

1. Start with \( y = \frac{1}{x - 5} \).
2. Swap \( x \) and \( y \): \( x = \frac{1}{y - 5} \).
3. Solve for \( y \):
\[
x(y - 5) = 1 \implies xy - 5x = 1 \implies xy = 1 + 5x \implies y = \frac{1 + 5x}{x}
\]
4. Replace \( y \) with \( f^{-1}(x) \):
\[
f^{-1}(x) = \frac{1 + 5x}{x}
\]
Note: The domain of \( f(x) \) is \( x \in (5, +\infty) \), so the range of \( f(x) \) is \( y \in (0, +\infty) \). Therefore, the domain of \( f^{-1}(x) \) is \( x \in (0, +\infty) \).

Answer:
\[
\boxed{f^{-1}(x) = \frac{1 + 5x}{x}, \, x \in (0, +\infty)}
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
&\text{(a)} \, f^{-1}(x) = \frac{x - 1}{4} \\
&\text{(b)} \, f^{-1}(x) = \frac{-2x - 1}{x - 3} \\
&\text{(c)} \, f^{-1}(x) = x^2 + 1, \, x \geq 0 \\
&\text{(d)} \, f^{-1}(x) = \frac{2x + 1}{x - 1}, \, x \in (1, +\infty) \\
&\text{(e)} \, f^{-1}(x) = \frac{1 + 5x}{x}, \, x \in (0, +\infty)
\end{aligned}
}
\]