Restrict Domain to
Make Invertible
Learn how to transform non-invertible functions into invertible ones by strategically restricting the domain.
Algebra 2
What is an Invertible Function?
A function \( f \) is invertible if there exists a function \( f^{-1} \) such that \( f^{-1}(f(x)) = x \) for every \( x \) in the domain, and \( f(f^{-1}(y)) = y \) for every \( y \) in the range.
The Horizontal Line Test
Why Does Invertibility Matter?
Solving Equations
To "undo" \( f(x) = y \) and solve for \( x \), you need \( f^{-1} \). Without it, solutions are ambiguous.
Inverse Trig Functions
\( \sin x, \cos x, \tan x \) are all non-invertible on their natural domains. \( \arcsin x, \arccos x, \arctan x \) exist only because we restrict.
Real-World Models
A height-vs-time projectile model isn't one-to-one. Restricting to the "going up" half lets you invert for time.
The Domain Restriction Theorem
In practice: find a largest interval where the function is either entirely going up or entirely going down. On that interval, the function passes the Horizontal Line Test and has an inverse.
Finding Monotonic Intervals
Each increasing or decreasing interval is a valid domain restriction. The function is one-to-one on any single monotonic piece.
One-to-One vs Not One-to-One
| One-to-One (Invertible) | Not One-to-One | |
|---|---|---|
| Definition | Each \( y \) has at most one \( x \) | Some \( y \) has multiple \( x \) values |
| HLT Result | Passes | Fails |
| Inverse exists? | Yes | No (unless domain restricted) |
| Graph shape | Strictly monotonic on domain | Turns back (has peaks/valleys) |
| Examples | \( f(x) = 2x + 3 \), \( e^x \), \( \ln x \) | \( x^2 \), \( \sin x \), \( |x| \) |
How to Restrict and Invert
1
Identify the vertex or turning point(s). Find where the function changes from increasing to decreasing (or vice versa).
2
Choose a monotonic interval. Pick the piece that is entirely increasing or entirely decreasing. The standard choice usually includes the vertex.
3
Write the restricted function. State the same rule but with the restricted domain, e.g., \( f(x) = x^2,\; x \geq 0 \).
4
Find the inverse algebraically. Swap \( x \) and \( y \), solve for \( y \), and state the domain of \( f^{-1} \).
5
Verify. Check that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) on the restricted domain.
Example 1: \( f(x) = x^2 \)
\( f \) and \( f^{-1} \) are Reflections
Example 2: \( f(x) = (x-3)^2 + 1 \)
Option A: \( x \leq 3 \)
Decreasing. Inverse: \( f^{-1}(x) = 3 - \sqrt{x - 1} \), domain \( x \geq 1 \).
Option B: \( x \geq 3 \)
Increasing. Inverse: \( f^{-1}(x) = 3 + \sqrt{x - 1} \), domain \( x \geq 1 \).
Finding the Inverse Algebraically
Choose the restriction \( x \geq 3 \). Now find \( f^{-1} \):
1
Write \( y = (x-3)^2 + 1 \)
2
Swap variables: \( x = (y-3)^2 + 1 \)
3
Isolate the squared term: \( x - 1 = (y-3)^2 \)
4
Take the square root: \( \sqrt{x - 1} = y - 3 \). We take the positive root because our restriction \( x \geq 3 \) means the inverse's range must be \( y \geq 3 \).
5
Solve: \( f^{-1}(x) = 3 + \sqrt{x - 1} \), domain: \( x \geq 1 \)
Example 3: \( f(x) = |x - 2| \)
Example 4: \( f(x) = \sin x \)
Common Mistakes to Avoid
✕ Forgetting to Restrict
Writing \( f^{-1}(x) = \sqrt{x} \) without stating \( x \geq 0 \) for the original function. The inverse only exists on the restricted domain.
✕ Wrong Root Sign
Choosing \( -\sqrt{x} \) when the restriction was \( x \geq 0 \), or \( +\sqrt{x} \) when the restriction was \( x \leq 0 \). The root sign must match the side of the vertex you kept.
✕ Ignoring the Vertex
Splitting the domain at \( x = 0 \) instead of at the vertex. For \( (x-3)^2 + 1 \), the split point is \( x = 3 \), not \( x = 0 \).
Summary of Common Restrictions
| Function | Restriction | Inverse |
|---|---|---|
| \( x^2 \) | \( x \geq 0 \) | \( \sqrt{x} \) |
| \( x^2 \) | \( x \leq 0 \) | \( -\sqrt{x} \) |
| \( (x-h)^2 + k \) | \( x \geq h \) | \( h + \sqrt{x - k} \) |
| \( |x| \) | \( x \geq 0 \) | \( x \) (identity) |
| \( \sin x \) | \( \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \) | \( \arcsin x \) |
| \( \cos x \) | \( [0,\, \pi] \) | \( \arccos x \) |
Question 1 of 5
Which domain restriction makes \( f(x) = x^2 - 4 \) invertible?
The vertex of \( x^2 - 4 \) is at \( x = 0 \). Restricting to \( x \geq 0 \) keeps the increasing half, making the function one-to-one. The interval \( [-2, 2] \) still contains both sides of the vertex.
Question 2 of 5
If \( f(x) = (x + 2)^2 \) is restricted to \( x \geq -2 \), what is \( f^{-1}(x) \)?
Swap \( x \) and \( y \): \( x = (y+2)^2 \). Take the positive root (since the inverse's range must be \( y \geq -2 \)): \( \sqrt{x} = y + 2 \), so \( y = \sqrt{x} - 2 \). This gives \( f^{-1}(x) = \sqrt{x} - 2 \).
Question 3 of 5
The function \( g(x) = |x - 5| \) is restricted to \( x \leq 5 \). What is the domain of \( g^{-1} \)?
The domain of \( g^{-1} \) equals the range of the restricted \( g \). On \( x \leq 5 \), \( g(x) = 5 - x \geq 0 \), so the range is \( [0, \infty) \). Therefore the domain of \( g^{-1} \) is \( x \geq 0 \).
Question 4 of 5
Why can't we define \( \arcsin x \) using \( \sin x \) on the interval \( [0, \pi] \)?
On \( [0, \pi] \), \( \sin x \) increases from \( 0 \) to \( 1 \), then decreases back to \( 0 \). It is not monotonic, so it fails the Horizontal Line Test. For example, \( \sin(\pi/6) = \sin(5\pi/6) = 0.5 \). The standard restriction \( [-\pi/2, \pi/2] \) keeps \( \sin x \) strictly increasing.
Question 5 of 5
If \( h(x) = (x - 1)^2 + 3 \) is restricted to \( x \leq 1 \), what is \( h^{-1}(7) \)?
Since \( x \leq 1 \), the inverse is \( h^{-1}(x) = 1 - \sqrt{x - 3} \). Plug in \( x = 7 \): \( h^{-1}(7) = 1 - \sqrt{7 - 3} = 1 - \sqrt{4} = 1 - 2 = -1 \). We use the negative root because we restricted to the left side of the vertex.