Inverse Functions (harder) Worksheet (with solutions) - Free Printable

Let's solve each of these inverse function problems step by step. The goal is to find the inverse function $ f^{-1}(x) $ for each given function $ f(x) $. To do this, we:

1. Replace $ f(x) $ with $ y $.
2. Swap $ x $ and $ y $.
3. Solve for $ y $.
4. Replace $ y $ with $ f^{-1}(x) $.

We'll go through each part carefully.

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

Step 1: Let $ y = 4x + 1 $

Step 2: Swap $ x $ and $ y $:
$ x = 4y + 1 $

Step 3: Solve for $ y $:
$ x - 1 = 4y $
$ y = \frac{x - 1}{4} $

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

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

Step 1: Let $ y = \frac{3x - 1}{x + 2} $

Step 2: Swap $ x $ and $ y $:
$ x = \frac{3y - 1}{y + 2} $

Step 3: Solve for $ y $:
Multiply both sides by $ y + 2 $:
$ x(y + 2) = 3y - 1 $
$ xy + 2x = 3y - 1 $

Bring all terms to one side:
$ xy - 3y + 2x + 1 = 0 $
Factor $ y $:
$ y(x - 3) = -2x - 1 $

Now solve for $ y $:
$ y = \frac{-2x - 1}{x - 3} $

We can write it as:
$$
f^{-1}(x) = \frac{-2x - 1}{x - 3}
$$

Alternatively, factor out a negative sign:
$$
f^{-1}(x) = \frac{-(2x + 1)}{x - 3}
$$

But the first form is fine.

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

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

Note: Domain of $ f(x) $ is $ x \geq 1 $, and range is $ y \geq 0 $.

Step 1: $ y = \sqrt{x - 1} $

Step 2: Swap $ x $ and $ y $:
$ x = \sqrt{y - 1} $

Step 3: Solve for $ y $:
Square both sides:
$ x^2 = y - 1 $
$ y = x^2 + 1 $

But remember: since $ f(x) = \sqrt{x - 1} $ outputs only non-negative values, the domain of $ f^{-1}(x) $ must be $ x \geq 0 $. So we restrict the domain of the inverse accordingly.

✔ Answer:
$$
f^{-1}(x) = x^2 + 1, \quad x \geq 0
$$

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

This is a rational function. We are told $ x > 2 $, so we’ll keep that in mind for domain/range considerations.

Step 1: $ y = \frac{x + 1}{x - 2} $

Step 2: Swap $ x $ and $ y $:
$ x = \frac{y + 1}{y - 2} $

Step 3: Solve for $ y $:
Multiply both sides by $ y - 2 $:
$ x(y - 2) = y + 1 $
$ xy - 2x = y + 1 $

Bring terms with $ y $ to one side:
$ xy - y = 2x + 1 $
$ y(x - 1) = 2x + 1 $

Now solve:
$ y = \frac{2x + 1}{x - 1} $

Now check domain and range.

Original function: $ x > 2 $. As $ x \to 2^+ $, $ f(x) \to \frac{3}{0^+} \to +\infty $. As $ x \to \infty $, $ f(x) \to 1^+ $. So range is $ (1, \infty) $.

So inverse function should have domain $ x > 1 $, and output $ y > 2 $.

Check: $ f^{-1}(x) = \frac{2x + 1}{x - 1} $, domain $ x > 1 $, and let’s verify:

As $ x \to 1^+ $, $ f^{-1}(x) \to \infty $; as $ x \to \infty $, $ f^{-1}(x) \to 2^+ $. So yes, output is $ > 2 $, matches original domain.

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

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

Domain: $ x > 5 $. As $ x \to 5^+ $, $ f(x) \to +\infty $. As $ x \to \infty $, $ f(x) \to 0^+ $. So range is $ (0, \infty) $.

Step 1: $ y = \frac{1}{x - 5} $

Step 2: Swap $ x $ and $ y $:
$ x = \frac{1}{y - 5} $

Step 3: Solve for $ y $:
Take reciprocal:
$ \frac{1}{x} = y - 5 $
$ y = \frac{1}{x} + 5 $

Now, since original function has range $ (0, \infty) $, the inverse function has domain $ x > 0 $.

Also, $ f^{-1}(x) = \frac{1}{x} + 5 $, and when $ x > 0 $, $ f^{-1}(x) > 5 $, which matches the original domain.

✔ Answer:
$$
f^{-1}(x) = \frac{1}{x} + 5, \quad x > 0
$$

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✔ Final Answers:

(a) $ f^{-1}(x) = \frac{x - 1}{4} $

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

(c) $ f^{-1}(x) = x^2 + 1, \quad x \geq 0 $

(d) $ f^{-1}(x) = \frac{2x + 1}{x - 1}, \quad x > 1 $

(e) $ f^{-1}(x) = \frac{1}{x} + 5, \quad x > 0 $

Let me know if you'd like graphical interpretations or verification steps!