Which Pair of Functions Are Inverses?
Understanding whether two functions undo each other’s effect is a core concept in algebra, calculus, and many applied fields. When you can show that applying one function after the other returns the original input, you have identified an inverse pair. This article explains the definition, the algebraic test, graphical clues, and practical examples to help you confidently decide which pair of functions are inverses.
1. What Does It Mean for Two Functions to Be Inverses?
Two functions, (f) and (g), are called inverse functions if they satisfy both of the following conditions for every (x) in the domains where the compositions make sense:
[ f(g(x)) = x \quad \text{and} \quad g(f(x)) = x ]
In words: plugging the output of (g) into (f) returns the original (x), and vice‑versa. When this holds, we write (g = f^{-1}) and (f = g^{-1}). The notation (f^{-1}) does not mean “(1/f)”; it denotes the function that reverses the action of (f).
2. Algebraic Test: Checking the Compositions
The most reliable way to verify an inverse pair is to compute the two compositions. Follow these steps:
- Write down the formulas for both functions, clearly indicating the variable (usually (x)).
- Substitute the inner function into the outer one: compute (f(g(x))). Simplify as much as possible.
- Repeat the process for the opposite order: compute (g(f(x))).
- Compare each result to the identity function (x). If both simplify exactly to (x) (over the shared domain), the pair is inverse.
Example 1: Linear Functions
Let (f(x) = 3x - 5) and (g(x) = \frac{x+5}{3}).
- (f(g(x)) = 3\left(\frac{x+5}{3}\right) - 5 = x + 5 - 5 = x)
- (g(f(x)) = \frac{(3x-5)+5}{3} = \frac{3x}{3} = x)
Both compositions give (x); therefore, (f) and (g) are inverses.
Example 2: Quadratic and Square‑Root Functions (with Domain Restriction)
Consider (f(x) = x^2) for (x \ge 0) and (g(x) = \sqrt{x}).
- (f(g(x)) = (\sqrt{x})^2 = x) (valid for (x \ge 0))
- (g(f(x)) = \sqrt{x^2} = |x|). Because we restricted (f) to non‑negative inputs, (|x| = x) for (x \ge 0).
Thus, with the domain (x \ge 0), the pair is inverse. Without the restriction, (g(f(x))) would fail for negative (x).
3. Graphical Method: Symmetry About the Line (y = x)
If you can sketch (or imagine) the graphs of two functions, they are inverses exactly when each graph is the mirror image of the other across the line (y = x). This visual test works because swapping (x) and (y) in the equation (y = f(x)) yields the inverse relation (x = f(y)) or (y = f^{-1}(x)) Easy to understand, harder to ignore. No workaround needed..
How to use it:
- Plot (y = f(x)).
- Plot the line (y = x) (a 45° diagonal through the origin).
- Reflect the graph of (f) across this line; the reflected curve should coincide with the graph of (g).
If the reflected image does not match (g), the functions are not inverses.
Note: This method is quick for simple functions but can be misleading if the graphs are not drawn accurately or if domain restrictions are ignored.
4. Common Pitfalls and How to Avoid Them
| Pitfall | Why It Happens | How to Avoid |
|---|---|---|
| Ignoring domain restrictions | A function may not be one‑to‑one over its natural domain, so its “inverse” fails the composition test. Consider this: | Always state the domain (and range) explicitly. Restrict the original function to make it one‑to‑one if needed. So |
| Confusing (f^{-1}) with (\frac{1}{f}) | The superscript (-1) looks like an exponent. In real terms, | Remember that (f^{-1}) denotes the inverse function, not the reciprocal. |
| Only checking one composition | Sometimes (f(g(x)) = x) but (g(f(x)) \neq x) (or vice‑versa). | Verify both compositions; both must equal (x). |
| Overlooking piecewise definitions | Piecewise functions can behave differently on different intervals. Think about it: | Treat each piece separately, ensuring the inverse matches the correct piece on the appropriate interval. Which means |
| Assuming symmetry guarantees inverses | A graph may appear symmetric about (y = x) due to scaling errors or limited viewing window. | Confirm algebraically; use the graph as a supportive check, not the sole proof. |
5. Step‑by‑Step Procedure to Determine If a Given Pair Is Inverse
Below is a concise checklist you can follow for any pair of functions (f) and (g).
- Identify the formulas and note any explicit domain restrictions.
- Compute (f(g(x))):
- Replace every (x) in (f) with the expression for (g(x)).
- Simplify using algebraic rules (distribute, combine like terms, factor, etc.).
- Compute (g(f(x))):
- Replace every (x) in (g) with the expression for (f(x)).
- Simplify similarly.
- Check each result:
- Does it reduce exactly to the variable (x)?
- Are there any leftover terms that depend on (x) in a non‑trivial way?
- State the conclusion:
- If both compositions equal (x) (over the shared domain), the pair is inverse.
- Otherwise, they are not inverses.
Tip: Keep a notebook of common inverse pairs (linear, exponential/logarithmic, trigonometric/inverse trigonometric) to speed up recognition.
6. Worked Examples
Example 3: Exponential and Logarithmic Functions
Let (f(x) = e^{x}) and (g(x) = \ln(x)) (natural log), with the domain of (g) restricted to (x > 0).
- (f(g(x)) = e^{\ln(x)} = x) for (x > 0).
- (g(f(x)) = \ln(e^{x}) = x) for all real (x).
Both compositions give (x); thus, (f) and (g) are inverse functions on the domain (x > 0) for (g) (which matches the range of (f)).
Example 4: Trigonometric Pair (Sine and
Example 4 (continued): Trigonometric Pair – Sine and Arcsine
Let
[ f(x)=\sin x ,\qquad g(x)=\arcsin x , ]
where we restrict the domain of (f) to (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]) so that it becomes one‑to‑one, and consequently the domain of (g) is ([-1,1]) with range (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]) Worth knowing..
-
Compute (f(g(x))):
[ f(g(x))=\sin(\arcsin x)=x \quad\text{for every }x\in[-1,1]. ]
-
Compute (g(f(x))):
[ g(f(x))=\arcsin(\sin x)=x \quad\text{provided }x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. ]
Both compositions return the input variable on their respective domains, confirming that (\sin) and (\arcsin) are inverse functions when the sine is restricted to the principal interval Worth keeping that in mind..
Example 5: Tangent and Arctangent with a Shift
Consider
[ f(x)=\tan!\left(x-\frac{\pi}{4}\right),\qquad g(x)=\arctan x+\frac{\pi}{4}. ]
Here we restrict (f) to the interval (\left(0,\frac{\pi}{2}\right)) (so that its argument stays within (\left(-\frac{\pi}{2},\frac{\pi}{2}\right))), making (f) one‑to‑one. The corresponding domain of (g) is all real numbers, and its range is (\left(0,\frac{\pi}{2}\right)).
-
(f(g(x))):
[ f(g(x))=\tan!\left(\bigl(\arctan x+\frac{\pi}{4}\bigr)-\frac{\pi}{4}\right) =\tan(\arctan x)=x \quad\text{for all }x\in\mathbb{R}. ]
-
(g(f(x))):
[ g(f(x))=\arctan!Which means \left(\tan! \left(x-\frac{\pi}{4}\right)\right)+\frac{\pi}{4} =\left(x-\frac{\pi}{4}\right)+\frac{\pi}{4}=x, ] valid for (x\in\left(0,\frac{\pi}{2}\right)) because the arctangent returns the principal value It's one of those things that adds up..
Thus (f) and (g) are inverse functions on the stated domains.
Quick Reference Table
| Function Pair | Typical Domain Restriction | Inverse Relationship |
|---|---|---|
| (e^{x}) , (\ln x) | (x>0) for (\ln) | (e^{\ln x}=x,;\ln(e^{x})=x) |
| (\sin x) , (\arcsin x) | (x\in[-\pi/2,\pi/2]) for (\sin) | (\sin(\arcsin x)=x,;\arcsin(\sin x)=x) |
| (\tan x) , (\arctan x) | (x\in(-\pi/2,\pi/2)) for (\tan) | (\tan(\arctan x)=x,;\arctan(\tan x)=x) |
| (ax+b) ( (a\neq0) ) , (\frac{x-b}{a}) | all real numbers | linear functions are always inverses after solving for (x) |
Conclusion
Verifying whether two functions are inverses hinges on a simple algebraic test: both compositions (f(g(x))) and (g(f(x))) must simplify to the identity function (x) on the overlap of their domains. Which means while graphical symmetry about the line (y=x) offers a helpful visual cue, it is not sufficient on its own—domain restrictions, piecewise behavior, and the distinction between the notation (f^{-1}) and the reciprocal (\frac{1}{f}) must all be considered carefully. So by following a systematic procedure—identifying formulas, computing both compositions, simplifying, and checking the results—you can confidently determine inverse relationships for a wide variety of functions, from elementary exponentials and logarithms to trigonometric and linear expressions. Mastery of this process not only prevents common mistakes but also deepens your understanding of how functions “undo” each other across mathematics.
