Inverse functions — finding f⁻¹(x), horizontal line test

Overview

An inverse function undoes what the original function does. If $f$ takes $x$ to $y$, then $f^{-1}$ takes $y$ back to $x$. Not every function has an inverse—only one-to-one (injective) functions do. The horizontal line test determines whether a function is one-to-one and thus invertible.

Connections:

• Function Composition — verifying inverses via $f(f^{-1}(x)) = x$
• Domain and Range — domain/range swap for inverses
• Bijective Functions — invertibility requires bijection
• Exponential and Logarithmic Functions — classic inverse pair
• Trigonometric Functions — restricting domains for invertibility

Core Intuition

For example, if $f(x) = 2x + 3$ takes $5\to 13$, then $f^{-1}(13) = 5$. The inverse "rewinds" the operation.

BUT: not every machine is reversible. If two different inputs give the same output (like $f(x) = x^2$ where $f(2) = f(-2) = 4$), you can't uniquely reverse it. That's why we need the one-to-one condition.

Definitions and Core Concepts

Key property: The graph of $f^{-1}$ is the reflection of $f$ across the line $y = x$.

Why this matters: Only one-to-one functions are invertible. If $f(2) = f(5) = 7$, then $f^{-1}(7)$ is ambiguous (is it 2 or 5?). No ambiguity means invertible.

Why it works: A horizontal line $y = c$ represents all points with the same $y$-value. If it hits the graph twice, two different $x$-values map to the same $y$, violating one-to-one.

Deriving the Inverse: The Swap Method

WHY this process works: If $(a, b)$ is on the graph of $f$ (meaning $f(a) = b$), then $(b, a)$ must be on the graph of $f^{-1}$ (meaning $f^{-1}(b) = a$). Swapping $x$ and $y$ in the equation literally swaps input and output roles.

Derivation from first principles:

• We want $f^{-1}$ such that $f(f^{-1}(x)) = x$
• Let $f^{-1}(x) = y$. Then $f(y) = x$ (by definition of inverse)
• So we need to solve $f(y) = x$ for $y$
• The swap method is just a notational trick to organize this solving process

Worked Examples

Step 1: Write $y = 3x - 7$

Step 2: Swap to get $x = 3y - 7$

Step 3: Solve for $y$: $x + 7 = 3y$ $y = \frac{x + 7}{3}$

Why this step? We're isolating $y$ to express the output in terms of the input.

Step 4: Therefore $f^{-1}(x) = \frac{x + 7}{3}$

Step 5: Verify: $f(f^{-1}(x)) = f\left(\frac{x+7}{3}\right) = 3 \cdot \frac{x+7}{3} - 7 = (x + 7) - 7 = x \,\checkmark$ $f^{-1}(f(x)) = f^{-1}(3x - 7) = \frac{(3x - 7) + 7}{3} = \frac{3x}{3} = x \,\checkmark$

Why verification matters: It's easy to make algebraic mistakes. Verification catches them.

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

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

Step 3: Solve for $y$: $x(y - 3) = 2y + 1$ $xy - 3x = 2y + 1$ $xy - 2y = 3x + 1$ $y(x - 2) = 3x + 1$ $y = \frac{3x + 1}{x - 2}$

Why this step? We collect all $y$ terms on one side to factor out $y$.

Step 4: Therefore $f^{-1}(x) = \frac{3x + 1}{x - 2}$, $x \neq 2$

Domain note: The denominator restriction changes because roles swap. Original's range limitation ($y \neq 2$ from the horizontal asymptote) becomes the inverse's domain restriction.

Verification (abbreviated): $f(f^{-1}(x)) = f\left(\frac{3x+1}{x-2}\right) = \frac{2 \cdot \frac{3x+1}{x-2} + 1}{\frac{3x+1}{x-2} - 3} = \frac{\frac{6x+2+x-2}{x-2}}{\frac{3x+1-3x+6}{x-2}} = \frac{7x}{7} = x \,\checkmark$

Horizontal Line Test: Draw $y = 4$. It intersects the parabola at $x = -2$ and $x = 2$. Two intersections means not one-to-one.

Why no inverse: If we tried to find $f^{-1}(4)$, we'd get two answers: $-2$ and $2$. An inverse must give exactly one output per input.

HOWEVER: If we restrict the domain to $x \geq 0$, then $f(x) = x^2$ becomes one-to-one on $[0, \infty)$. Now we can find the inverse:

• $y = x^2$, $x \geq 0$
• $x = y^2$, $y \geq 0$
• $y = \sqrt{x}$ (positive root only)
• $f^{-1}(x) = \sqrt{x}$

Why restrict? Domain restriction removes the ambiguity. This technique is crucial for trig functions.

Step 1: $y = \sqrt[3]{x + 5} - 2$

Step 2: $x = \sqrt[3]{y + 5} - 2$

Step 3: Solve for $y$: $x + 2 = \sqrt[3]{y + 5}$ $(x + 2)^3 = y + 5$ $y = (x + 2)^3 - 5$

Why cube both sides? Cubing undoes the cube root, just as the inverse operation should.

Step 4: $f^{-1}(x) = (x + 2)^3 - 5$

Domain/Range: Since cube root is defined for all real numbers and is one-to-one, both $f$ and $f^{-1}$ have domain and range $\mathbb{R}$.

Common Mistakes & How to Fix Them

Why it feels right: The $-1$ exponent notation is borrowed from reciprocals ($x^{-1} = 1/x$).

The truth: $f^{-1}$ is inverse function, NOT reciprocal.

• $f^{-1}(f(x)) = x$ (composition gives identity)
• $\frac{1}{f(x)} \cdot f(x) = 1$ (multiplication gives 1)

Example: For $f(x) = 2x$:

• Inverse: $f^{-1}(x) = \frac{x}{2}$, and $f(f^{-1}(x)) = 2 \cdot \frac{x}{2} = x$ ✓
• Reciprocal: $\frac{1}{f(x)} = \frac{1}{2x}$, which is completely different

Fix: Always verify your inverse using composition, never by multiplication.

Why it feels right: The algebra is correct.

The truth: $f(x) = \sqrt{x - 2}$ has domain $x \geq 2$ and range $y \geq 0$. When we swap, $f^{-1}$ must have domain $x \geq 0$ (the original range) and range $y \geq 2$ (the original domain).

So $f^{-1}(x) = x^2 + 2$ only for $x \geq 0$.

Fix: Track domain/range swaps explicitly:

• Original domain → inverse range
• Original range → inverse domain

Why it feels right: The algorithm is mechanical and always produces something.

The truth: If $f$ isn't one-to-one, the "inverse" you find won't actually work. For $f(x) = x^2$ (all reals):

• Swap method gives $f^{-1}(x) = \pm\sqrt{x}$
• But this isn't a function (two outputs for one input)!

Fix: Always apply horizontal line test first. If it fails, restrict the domain before finding inverse.

Horizontal Line Test — Geometric Proof

WHY does the horizontal line test work?

Claim: $f$ is one-to-one $\iff$ every horizontal line intersects the graph at most once.

Proof ($\Rightarrow$): Suppose $f$ is one-to-one. Assume for contradiction that a horizontal line $y = c$ intersects the graph at two points $(x_1, c)$ and $(x_2, c)$ with $x_1 \neq x_2$. Then $f(x_1) = c$ and $f(x_2) = c$, so $f(x_1) = f(x_2)$ but $x_1 \neq x_2$. This contradicts one-to-one. ✓

Proof ($\Leftarrow$): Suppose every horizontal line intersects at most once. To show one-to-one, assume $f(x_1) = f(x_2) = c$. Then $(x_1, c)$ and $(x_2, c)$ are both on the graph, so they lie on the horizontal line $y = c$. Since this line intersects at most once, we must have $x_1 = x_2$. Thus $f$ is one-to-one. ✓

Domain and Range of Inverses

Derivation:

• Let $y \in \text{Range}(f)$. Then $y = f(x)$ for some $x \in A$. By definition of inverse, $f^{-1}(y) = x$, so $y \in \text{Domain}(f^{-1})$.
• Conversely, if $y \in \text{Domain}(f^{-1})$, then $f^{-1}(y)$ is defined, meaning there exists $x$ such that $f(x) = y$, so $y \in \text{Range}(f)$.

Why this matters: When finding inverses algebraically, you must transfer domain restrictions correctly.

Properties of Inverse Functions

2. Inverse of Inverse: $(f^{-1})^{-1} = f$ Derivation: If $g = f^{-1}$, then $f(g(x)) = x$ and $g(f(x)) = x$. This means $f$ satisfies the definition of the inverse of $g$, so $g^{-1} = f$.

3. Composition Property: $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$

   Derivation: Toundo "do $g$ then do $f$", we must "undo $f$ then undo $g$" (reverse order). Formally: $(g^{-1} \circ f^{-1})(f \circ g)(x) = g^{-1}(f^{-1}(f(g(x))) = g^{-1}(g(x)) = x$

Active Recall Questions

#flashcards/maths

What is the definition of an inverse function? :: A function $f^{-1}$ such that $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$. It "undoes" what $f$ does.

State the Horizontal Line Test :: A function is one-to-one if and only if every horizontal line intersects its graph at most once.

An inverse function is like having a decoder machine. You give it the coded number (13), and it tells you the original number (5). The decoder does the opposite operations in reverse order: first subtract 3, then divide by 2.

But here's the catch: not every code machine has a decoder! If your machine sometimes gives the same code for different numbers (like squaring: both 3 and -3 give 9), then you can't decode uniquely. You wouldn't know if 9 came from 3 or -3.

The horizontal line test is a quick way to check: draw your machine's graph, then draw horizontal lines. If any line touches the graph twice, your machine isn't decodable (two different inputs gave the same output).

To find the decoder formula: write down your machine's rule, swap the input and output, then solve to get the input by itself. That's your decoder!

For horizontal line test: "Horizontal = One-to-One" — if a horizontal line hits once, the function is one-to-one.

Summary

To find an inverse function $f^{-1}(x)$:

1. Verify $f$ is one-to-one (use horizontal line test)
2. Use the swap method: $y = f(x) \to x = f(y) \to$ solve for $y$
3. Remember domain and range swap
4. Always verify using composition

The inverse "undoes" the function. Graphically, it's a reflection across $y = x$. Only one-to-one functions are invertible—the horizontal line test checks this condition by ensuring each output comes from exactly one input.

Concept Map

Hinglish (regional understanding)