Inverse Functions
Introduced| English |
|---|
| inverse/ɪnˈvɜːs/ |
| domain/dəˈmeɪn/ |
| one-to-one/wʌn tuː wʌn/ |
| reflection/rɪˈflekʃn/ |
Undoing a function
- Every function does something to an input; its inverse undoes it.
- Add 3, and the inverse subtracts 3. Double it, and the inverse halves.
- The inverse runs the machine backward, turning outputs into inputs.
- This "undo" is exactly how we solve equations built from functions.
What an inverse does
- The inverse 反函数 of $f$, written $f^{-1}$, reverses every input-output pair.
- If $f$ sends $2 \to 5$, then $f^{-1}$ sends $5 \to 2$.
- So the domain of $f$ becomes the range of $f^{-1}$, and vice versa.
- Composing them undoes everything: $f^{-1}(f(x)) = x$.
The inverse of a function…
If $f$ sends $2 \to 5$, then $f^{-1}$ sends $5 \to 2$: the inverse undoes what the function did.
It must be one-to-one
- An inverse function exists only if $f$ is one-to-one 一一对应.
- One-to-one means each output comes from exactly one input.
- If two inputs share an output, running backward is ambiguous — no single inverse.
- The horizontal-line test checks this: no horizontal line may hit the graph twice.
Reflect a line across y = x to see its inverse
y = ax + b
This line is f(x) = 2x + 1. Its inverse swaps x and y, giving the mirror image across the line y = x.
A function has an inverse function only if it is ____ (each output comes from exactly one input).
If two inputs share an output, the reverse map is ambiguous — so only a one-to-one function is invertible.
Finding the inverse rule
- Write $y = f(x)$, then swap $x$ and $y$.
- Now solve the new equation for $y$ — that expression is $f^{-1}(x)$.
- For $f(x) = 2x + 1$: swap to $x = 2y + 1$, solve → $y = \dfrac{x-1}{2}$.
- Swapping reverses the roles of input and output, which is the whole idea.
To find the rule for $f^{-1}$, you…
Swapping $x$ and $y$ reverses the roles of input and output; solving for $y$ gives the inverse rule.
Select all true statements about inverse functions.
Not every function is invertible — it must be one-to-one first. The other three are correct.
Reflection across y = x
- Swapping $(a,b)$ to $(b,a)$ mirrors every point across the line $y = x$.
- So the graph of $f^{-1}$ is the reflection 反射 of the graph of $f$ in that line.
- Wherever $f$ crosses the y-axis, $f^{-1}$ crosses the x-axis, and so on.
- You can verify an inverse by checking $f(f^{-1}(x)) = x$.
The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$.
Swapping every $(a,b)$ to $(b,a)$ mirrors the whole graph across $y = x$.
The notation $f^{-1}$ means the inverse function, not the reciprocal $\dfrac{1}{f}$. Despite the "$-1$", $f^{-1}(x)$ is never $\dfrac{1}{f(x)}$ — it is the function that undoes $f$.
Find the inverse of $f(x) = 3x - 6$.
- Write $y = 3x - 6$, then swap: $x = 3y - 6$.
- Solve for $y$: $x + 6 = 3y$, so $y = \dfrac{x+6}{3}$.
- Check: $f^{-1}(f(2)) = f^{-1}(0) = 2$ — it undoes $f$. ✓
An inverse function $f^{-1}$ reverses each input-output pair, so it undoes $f$. It exists only when $f$ is one-to-one. Find it by swapping $x$ and $y$ and solving; its graph is the reflection of $f$ across the line $y = x$.