Inverse Functions

🔄 Inverse Functions

Part 1 of 5 — The Idea of "Undoing"

Topics in This Part

🔑 Key Concept: An inverse function reverses what the original function did. If $f$ turns $3$ into $11$, then $f^{-1}$ turns $11$ back into $3$. The inverse runs the machine backwards.

A Function as a Machine

Think of a function as a machine that takes an input, does something to it, and spits out an output.

$f(x) = x + 5 \quad\text{adds 5}$

The inverse machine undoes that step — it subtracts 5:

$f^{-1}(x) = x - 5$

Stacked Operations Undo in Reverse Order

If $f$ multiplies by $2$, then adds $3$, the inverse must first subtract $3$, then divide by $2$ — opposite operations in the opposite order.

💡 Like getting dressed: socks-then-shoes undoes as shoes-off-then-socks-off. Last on, first off.

Concept Check 🎯

Swapping Inputs and Outputs

Every ordered pair tells you an input and an output. The inverse simply swaps them:

$\text{if } (a, b) \text{ is on } f, \quad\text{then } (b, a) \text{ is on } f^{-1}.$

Example

If $f = \{(1, 4),\ (2, 7),\ (3, 10)\}$, then

$f^{-1} = \{(4, 1),\ (7, 2),\ (10, 3)\}.$

🔑 Domain ↔ Range Swap: The domain of $f$ becomes the range of $f^{-1}$, and the range of $f$ becomes the domain of $f^{-1}$. The inputs and outputs trade places.

Swap the Coordinates 🧮

A point on $f$ is given. Enter the matching point on $f^{-1}$ as the two coordinates.

1) $(6, 9)$ is on $f$. On $f^{-1}$: $(\,?,\ ?\,)$ — enter the $x$-coordinate, then the $y$-coordinate. 2) $(-2, 5)$ is on $f$. On $f^{-1}$: $(\,?,\ ?\,)$ — enter the $x$-coordinate, then the $y$-coordinate.

A Warning About the Notation

The symbol $f^{-1}$ is read "f inverse." It is not an exponent.

$f^{-1}(x) \ne \frac{1}{f(x)}$

⚠️ Common trap: $f^{-1}(x)$ does not mean $\dfrac{1}{f(x)}$. The $-1$ here is a label that means "inverse function," not a power. (The reciprocal $\frac{1}{f(x)}$ is a completely different thing.)

In Part 2 we'll learn the algebra trick that finds $f^{-1}(x)$ from a formula — and it's just the coordinate swap written with $x$ and $y$.

Reading the Notation 🔽

Pick the statement that is true in each row.

🔄 Inverse Functions

Part 2 of 5 — Finding the Inverse Algebraically

🔑 The Three-Step Recipe: (1) Write $y = f(x)$. (2) Swap $x$ and $y$. (3) Solve the new equation for $y$. That solved-for-$y$ is $f^{-1}(x)$.

The Swap-and-Solve Method

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

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

Worked Example: $f(x) = 2x + 3$

$y = 2x + 3 \quad\xrightarrow{\text{swap}}\quad x = 2y + 3$

Now solve for $y$:

$x - 3 = 2y \quad\Rightarrow\quad y = \frac{x - 3}{2}$

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

✅ Quick check: $f(2) = 2(2)+3 = 7$, and $f^{-1}(7) = \dfrac{7-3}{2} = 2$. The inverse sent $7$ back to $2$. ✓

Worked Example: $f(x) = \dfrac{x - 5}{3}$

$y = \frac{x - 5}{3} \quad\xrightarrow{\text{swap}}\quad x = \frac{y - 5}{3}$

Multiply both sides by $3$, then add $5$:

$3x = y - 5 \quad\Rightarrow\quad y = 3x + 5$

$\boxed{f^{-1}(x) = 3x + 5}$

Worked Example: $f(x) = x^3 + 1$

$y = x^3 + 1 \quad\xrightarrow{\text{swap}}\quad x = y^3 + 1$

$y^3 = x - 1 \quad\Rightarrow\quad y = \sqrt[3]{x - 1}$

$\boxed{f^{-1}(x) = \sqrt[3]{x - 1}}$

💡 The cube root undoes the cube. Whatever the last operation was on $x$, you undo it first when solving.

Concept Check 🎯

Find the Inverse 🧮

Find $f^{-1}(x)$ and evaluate it at the given input. Enter the value of $f^{-1}$ at that input.

1) $f(x) = x + 8$. Find $f^{-1}(10) = \,?$ 2) $f(x) = 4x$. Find $f^{-1}(20) = \,?$ 3) $f(x) = 3x - 1$. Find $f^{-1}(11) = \,?$

Build the Inverse 🔽

You're inverting $f(x) = 2x + 3$. Choose the result at each stage.

🔄 Inverse Functions

Part 3 of 5 — Verifying Inverses with Composition

🔑 The Test: Two functions are inverses only if composing them in both orders gives back $x$: $f(f^{-1}(x)) = x \quad\text{and}\quad f^{-1}(f(x)) = x.$

Composition Returns the Input

If $f$ and $g$ are inverses, then feeding the output of one straight into the other lands you exactly where you started.

$f(g(x)) = x \qquad\text{and}\qquad g(f(x)) = x$

You must check both orders — passing only one is not enough to confirm a true inverse.

Worked Example: Are $f(x) = 2x + 3$ and $g(x) = \dfrac{x - 3}{2}$ inverses?

Order 1 — $f(g(x))$:

$f\!\left(\frac{x-3}{2}\right) = 2\cdot\frac{x-3}{2} + 3 = (x - 3) + 3 = x \ ✓$

Order 2 — $g(f(x))$:

$g(2x + 3) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x \ ✓$

✅ Both compositions return $x$, so $f$ and $g$ are inverses.

Concept Check 🎯

Compose to Verify 🧮

Let $f(x) = 3x - 6$ and $g(x) = \dfrac{x + 6}{3}$ (its candidate inverse).

1) Compute $f(g(0))$. First $g(0) = \,?$ 2) Then $f$ of that result: $f(g(0)) = \,?$ 3) Compute $g(f(2))$. (You should get the same value as the input, $2$.) Enter $g(f(2)) = \,?$

When Composition Fails

Example: Are $f(x) = x^2$ and $g(x) = \sqrt{x}$ inverses?

$f(g(x)) = (\sqrt{x})^2 = x \quad\text{(for } x \ge 0\text{)}$

$g(f(x)) = \sqrt{x^2} = |x|$

The second composition gives $|x|$, not $x$. For a negative input like $x = -3$:

$g(f(-3)) = \sqrt{(-3)^2} = \sqrt{9} = 3 \ne -3$

⚠️ So $f(x) = x^2$ is not invertible over all real numbers — two different inputs ($3$ and $-3$) share the same output ($9$), so the reverse direction is ambiguous. We fix this in Part 4 by restricting the domain.

Verify by Composition 🔽

Test whether $f(x) = x + 4$ and $g(x) = x - 4$ are inverses.

🔄 Inverse Functions

Part 4 of 5 — Graphs, One-to-One, and Domain Restrictions

🔑 The Mirror: The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$. That single picture explains the whole coordinate swap.

Reflection Across $y = x$

Because the inverse swaps each $(a, b)$ for $(b, a)$, its graph is the original flipped over the diagonal line $y = x$.

$(a, b) \text{ on } f \quad\Longleftrightarrow\quad (b, a) \text{ on } f^{-1}$

💡 Any point already on the line $y = x$ (like $(4, 4)$) stays put when reflected — it is its own mirror image. So $f$ and $f^{-1}$ always cross each other right on that line when they meet.

Concept Check 🎯

One-to-One: The Horizontal Line Test

A function has an inverse that is also a function only if it is one-to-one — every output comes from exactly one input.

• Vertical Line Test → checks whether a graph is a function (no input has two outputs).
• Horizontal Line Test → checks whether a function is one-to-one (no output has two inputs).

🔑 Horizontal Line Test: If every horizontal line crosses the graph at most once, the function is one-to-one and has an inverse function.

Why $f(x) = x^2$ fails

A horizontal line like $y = 9$ hits the parabola at both $x = 3$ and $x = -3$. Two inputs share one output, so it is not one-to-one — and so it has no inverse function over all reals.

Restricting the Domain

We can rescue a non-one-to-one function by restricting its domain to a piece that passes the horizontal line test.

Example: $f(x) = x^2,\ x \ge 0$

Keeping only the right half (where $x \ge 0$) makes it one-to-one. Now invert:

$y = x^2 \ (x \ge 0) \xrightarrow{\text{swap}} x = y^2 \Rightarrow y = \sqrt{x}$

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

The range of $f$ (here $y \ge 0$) becomes the domain of $f^{-1}$, so $f^{-1}(x) = \sqrt{x}$ is defined for $x \ge 0$.

Example: $f(x) = \sqrt{x - 4}$

$y = \sqrt{x - 4} \xrightarrow{\text{swap}} x = \sqrt{y - 4} \Rightarrow x^2 = y - 4 \Rightarrow y = x^2 + 4$

Because the original output $\sqrt{x-4}$ is never negative, $f^{-1}(x) = x^2 + 4$ carries the restriction $x \ge 0$.

⚠️ Always carry the domain restriction along — without "$x \ge 0$," the formula $x^2 + 4$ would not be a true inverse.

Domain-Restricted Inverses 🧮

1) $f(x) = x^2 + 1$ with $x \ge 0$. Its inverse is $f^{-1}(x) = \sqrt{x - c}$. Enter $c = \,?$ 2) Using that inverse, compute $f^{-1}(10) = \,?$ 3) A one-to-one function $f$ has range $y \ge 3$. What is the smallest $x$-value in the domain of $f^{-1}$? Enter it.

One-to-One Reasoning 🔽

🔄 Inverse Functions

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) describe an inverse as an "undo," (2) find $f^{-1}$ by swap-and-solve, (3) verify with composition, and (4) handle graphs, one-to-one, and domain restrictions. Let's bring it all together.

Quick Reference

⚠️ Remember: $f^{-1}(x) \ne \dfrac{1}{f(x)}$, and always carry along any domain restriction.

Mixed Practice 🎯

Apply It 🧮

A temperature converter uses $f(C) = \dfrac{9}{5}C + 32$ to turn Celsius into Fahrenheit. The inverse $f^{-1}(F) = \dfrac{5}{9}(F - 32)$ converts back.

1) Convert $20$°C to Fahrenheit: $f(20) = \,?$ (°F) 2) Convert that result back: $f^{-1}(68) = \,?$ (°C) 3) Convert $212$°F to Celsius: $f^{-1}(212) = \,?$ (°C)

Exit Quiz ✅

Answer all three to finish the lesson.