Inverse Function Rules

Why This Matters

Inverse functions are one of those topics that shows up everywhere in Pre-Calculus—and they're not going away when you hit Calculus. You're being tested on your ability to understand how functions can be "undone," which connects directly to solving equations, working with logarithms and exponentials, and analyzing trigonometric relationships. The concept of reversibility is fundamental: if you can't find an inverse, you can't solve for the original input, and that limitation shapes everything from graphing to real-world applications.

What makes inverse functions tricky on exams isn't the basic definition—it's knowing when a function has an inverse, how to find it algebraically, and why certain functions need domain restrictions. You'll need to connect the horizontal line test to one-to-one functions, understand why graphs reflect over $y = x$, and recognize inverse pairs like exponentials and logarithms on sight. Don't just memorize the steps for finding an inverse—know what concept each rule illustrates and why it matters.

---

Foundational Concepts: What Makes Inverses Work

Before you can find or use inverse functions, you need to understand what they are and what conditions must be met. The core idea is reversibility—an inverse function undoes what the original function did.

Definition of an Inverse Function

• An inverse function reverses the original function's operation—if $f(x)$ takes input $x$ and produces output $y$, then $f^{-1}(y)$ takes $y$ back to $x$
• Notation matters: $f^{-1}$ means the inverse of $f$, not $\frac{1}{f(x)}$—this is a common exam trap
• The inverse exists only when the reversal is unambiguous—every output must trace back to exactly one input

One-to-One Function Requirement

• A function must be one-to-one (injective) to have an inverse—each output corresponds to exactly one input
• The horizontal line test determines this graphically: if any horizontal line crosses the graph more than once, the function fails
• Why it matters: functions like $f(x) = x^2$ aren't one-to-one on their natural domain, which is why we restrict them

Domain and Range Swap

• The domain of $f$ becomes the range of $f^{-1}$—inputs and outputs literally trade places
• The range of $f$ becomes the domain of $f^{-1}$—this swap is automatic and unavoidable
• Exam application: when asked for the domain of an inverse, look at the range of the original function

---

Graphical and Algebraic Relationships

Understanding how inverses appear on graphs and how to find them algebraically are the two most testable skills in this unit. The reflection property connects visual intuition to the algebraic process.

Reflection Over $y = x$

• The graph of $f^{-1}$ is the reflection of $f$ over the line $y = x$—this is the visual signature of inverse functions
• Coordinate swap: if $(a, b)$ is on the graph of $f$, then $(b, a)$ is on the graph of $f^{-1}$
• Use this to check your work—sketch both graphs and confirm the reflection relationship

Finding Inverses Algebraically

• Step 1: Replace $f(x)$ with $y$, then solve for $x$ in terms of $y$
• Step 2: Swap $x$ and $y$ to write the inverse function—this reflects the coordinate swap graphically
• Step 3: Verify by composition—check that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$

Composition Yields the Identity Function

• The defining property: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ for all valid inputs
• The identity function simply returns its input unchanged—composition with an inverse always produces this
• Verification tool: if composition doesn't give you $x$, you've made an error finding the inverse

---

Domain Restrictions and Invertibility

Sometimes functions aren't naturally one-to-one, but we can make them invertible by restricting their domain. This is how we create useful inverse functions for parabolas, trig functions, and more.

Restricting Domain to Create Invertible Functions

• Restrict the domain to an interval where the function is one-to-one—this "forces" the function to pass the horizontal line test
• Common example: $f(x) = x^2$ restricted to $x \geq 0$ has inverse $f^{-1}(x) = \sqrt{x}$
• The restriction choice matters—different restrictions yield different inverse functions with different domains

Inverse Trigonometric Functions

• Inverse trig functions find angles from ratios—$\sin^{-1}(x)$, $\cos^{-1}(x)$, and $\tan^{-1}(x)$ return angle measures
• Each has a restricted range to ensure one-to-one behavior: $\sin^{-1}$ outputs $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\cos^{-1}$ outputs $[0, \pi]$, $\tan^{-1}$ outputs $(-\frac{\pi}{2}, \frac{\pi}{2})$
• Domain restrictions on the original trig functions make these inverses possible—memorize the principal value ranges

---

Special Inverse Pairs

Some function pairs are so fundamental that you should recognize them instantly. Exponentials and logarithms are the classic example of inverse functions in action.

Logarithmic and Exponential Functions as Inverses

• $f(x) = a^x$ and $f^{-1}(x) = \log_a(x)$ are inverse functions—this relationship defines what logarithms mean
• Key identities: $a^{\log_a(x)} = x$ and $\log_a(a^x) = x$ follow directly from the composition property
• Base requirements: $a > 0$ and $a \neq 1$—otherwise the functions aren't well-defined or aren't one-to-one

Derivative of Inverse Functions

• The derivative formula: $(f^{-1})'(y) = \frac{1}{f'(x)}$ where $y = f(x)$
• Interpretation: the rate of change of the inverse is the reciprocal of the original function's rate of change
• Calculus preview: this connects inverse functions to implicit differentiation and the chain rule

---

Quick Reference Table

---

Self-Check Questions

1. What do the horizontal line test and the one-to-one requirement have in common, and why are both necessary for a function to have an inverse?

2. If $f(x)$ has domain $[-3, 5]$ and range $[0, 10]$, what are the domain and range of $f^{-1}(x)$?

3. Compare and contrast the process of verifying inverses graphically (reflection) versus algebraically (composition). When would you use each method?

4. Why does $f(x) = \sin(x)$ require a domain restriction to have an inverse, while $f(x) = e^x$ does not? What restriction is used for sine?

5. Given $f(x) = 2x + 5$, find $f^{-1}(x)$ algebraically and verify your answer using composition. What does the reflection property predict about the slopes of $f$ and $f^{-1}$?