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You know that $f^{-1}$ "undoes" $f$. But how do you actually find a formula for $f^{-1}$? The key insight: if $y = f(x)$ describes the forward process, then solving for $x$ in terms of $y$ describes the reverse. After that, we just swap variable names to match standard notation.
| Property | Value |
|---|---|
| Chapter | Chapter 6: Inverse Functions |
| Section | §6.1 Inverse Functions and Their Derivatives |
| Difficulty | Intermediate |
| Time | ~20 minutes |
$$\boxed{\text{To find } f^{-1}:}$$
| Step | Action | Why It Works |
|---|---|---|
| 1 | Write $y = f(x)$ | Sets up the equation relating input and output |
| 2 | Solve for $x$ in terms of $y$ | Reverses the process; now $x$ depends on $y$ |
| 3 | Interchange $x$ and $y$ | Standard convention: independent variable is $x$ |
The result is $y = f^{-1}(x)$.
After Step 2, you have $x$ written as a function of $y$. Mathematically, this is the inverse function: it tells you the input that produces a given output.
But conventionally, we like to write functions as "$y$ = something in $x$" with $x$ as the input. Swapping $x$ and $y$ is just a relabeling to match this convention. It doesn't change the relationship, only the names.
Find the inverse of $f(x) = 3x + 7$.
Step 1: Write $y = 3x + 7$
Step 2: Solve for $x$: $$y - 7 = 3x$$ $$x = \frac{y - 7}{3}$$
Step 3: Interchange $x$ and $y$: $$y = \frac{x - 7}{3}$$
Result: $f^{-1}(x) = \frac{x - 7}{3}$
Verification: Check that $f(f^{-1}(x)) = x$: $$f\left(\frac{x-7}{3}\right) = 3 \cdot \frac{x-7}{3} + 7 = (x-7) + 7 = x \quad ✓$$
Find the inverse of $f(x) = x^3 - 5$.
Step 1: Write $y = x^3 - 5$
Step 2: Solve for $x$: $$y + 5 = x^3$$ $$x = \sqrt[3]{y + 5}$$
Step 3: Interchange $x$ and $y$: $$y = \sqrt[3]{x + 5}$$
Result: $f^{-1}(x) = \sqrt[3]{x + 5}$
Find the inverse of $f(x) = \frac{x + 1}{x - 2}$ (for $x \neq 2$).
Step 1: Write $y = \frac{x + 1}{x - 2}$
Step 2: Solve for $x$ (this requires more algebra):
Multiply both sides by $(x - 2)$: $$y(x - 2) = x + 1$$
Expand: $$yx - 2y = x + 1$$
Collect $x$ terms on one side: $$yx - x = 2y + 1$$ $$x(y - 1) = 2y + 1$$ $$x = \frac{2y + 1}{y - 1}$$
Step 3: Interchange $x$ and $y$: $$y = \frac{2x + 1}{x - 1}$$
Result: $f^{-1}(x) = \frac{2x + 1}{x - 1}$ (for $x \neq 1$)
When the original function has a restricted domain (to ensure it's one-to-one), you must track what happens to that restriction.
Example: Find the inverse of $f(x) = x^2 + 4$ for $x \geq 0$.
Step 1: $y = x^2 + 4$
Step 2: $y - 4 = x^2$, so $x = \pm\sqrt{y - 4}$
Since the domain is $x \geq 0$, we take the positive root: $x = \sqrt{y - 4}$
Step 3: $y = \sqrt{x - 4}$
Result: $f^{-1}(x) = \sqrt{x - 4}$ with domain $x \geq 4$
| Mistake | Example | Correction |
|---|---|---|
| Forgetting to swap $x$ and $y$ | Getting $x = \frac{y-7}{3}$ and stopping | Continue: swap to get $y = \frac{x-7}{3}$ |
| Taking wrong root | $x^2 = 4 \Rightarrow x = 2$ | Check domain: might need $x = -2$ |
| Not checking domain | Finding inverse without noting restrictions | State domain of $f^{-1}$ (= range of $f$) |
Find the inverse of $f(x) = 5x - 7$.
Find the inverse of $g(x) = \sqrt{x + 3}$ (for $x \geq -3$).
Find the inverse of $h(x) = x^2 - 4x$ for $x \geq 2$.
Find the inverse of $f(x) = \frac{3x - 2}{x + 4}$ (for $x \neq -4$).
(a) Find the inverse of $f(x) = \frac{1 - x}{1 + x}$ (for $x \neq -1$).
(b) What do you notice? Explain why this happens geometrically.
(c) Find all values of $a$ such that $g(x) = \frac{a - x}{1 + ax}$ is self-inverse (meaning $g = g^{-1}$).
Question 1: If $f(x) = x^3 + 5$, what is the first step in finding $f^{-1}$?
(A) Take the cube root of $x$ (B) Write $y = x^3 + 5$ and solve for $x$ (C) Swap $f$ and $f^{-1}$ (D) Subtract 5 from $f(x)$
(B) The first step is to write $y = x^3 + 5$ and then solve for $x$ in terms of $y$. This gives $x = \sqrt[3]{y-5}$. Then swap $x$ and $y$ to get $f^{-1}(x) = \sqrt[3]{x-5}$.
Question 2: When finding the inverse of $f(x) = x^2$ for $x \leq 0$, you solve $y = x^2$ and get $x = \pm\sqrt{y}$. Why do you choose $x = -\sqrt{y}$?
(A) The negative is always correct for square roots (B) The domain restriction $x \leq 0$ requires the negative root (C) It doesn't matter which root you choose (D) You should always use the positive root
(B) The domain restriction $x \leq 0$ means we're only considering the left half of the parabola. When we solve for $x$, we must honor this restriction, so we take the negative root: $f^{-1}(x) = -\sqrt{x}$.
The "Unwrapping" Picture:
Think of a function as wrapping a gift: $f(x) = 3x + 7$ means "triple it, then add 7."
Finding the inverse is like writing instructions to unwrap: "subtract 7, then divide by 3."
The order reverses! Whatever $f$ did last, $f^{-1}$ undoes first.
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|---|---|---|
| Inverse Function Definition | Chapter 6 | Graphing Inverse Functions |
Last updated: 2026-01-23