√((−3)²) = √9 = 3, not −3. So √(x²) is |x|, and that one fact is why x² has no inverse until you restrict its domain.
On the right half they are the same function. On the left half they are not. A rule that holds on half the number line is the kind that survives until an exam.
√ means the non-negative square root, by definition. There has to be a choice, because 9 has two square roots and a function may only return one, and the choice made is the positive one.
So squaring and then rooting does not get you back:
√((−3)²) = √9 = 3
and in general √(x²) = |x|. The squaring destroyed the sign and the rooting cannot invent it back.
This is the same fact as "x² is not one-to-one", said arithmetically. 3 and −3 both go to 9, so an inverse would have to send 9 to both, and no function does that. √(x²) = |x| and "x² has no inverse" are one statement.
Restricting the domain is what rescues it. On x ≥ 0 the function x² is one-to-one, and then its inverse is √x and everything works: the inverse of 9 is 3, with no ambiguity, because −3 is no longer in the domain.
The restriction is not a technicality to mention and move past; it is the thing that makes the inverse exist, and a question that asks for an inverse of a quadratic always expects it to be stated. On x ≤ 0 the inverse would be −√x instead, which is the other half of the same answer.
Two symmetries, each a one-line test:
| Name | Test | Symmetry | Example |
|---|---|---|---|
| Even | f(−x) = f(x) | Reflection in the y-axis | x², cos x |
| Odd | f(−x) = −f(x) | Rotation of 180° about the origin | x³, sin x |
| Neither | both fail | none | x² + x |
Do the test on numbers and it takes seconds. For x³: f(−4) = −64 and −f(4) = −64, so it is odd. For x² + x: f(−2) = 2, f(2) = 6 and −f(2) = −6, so 2 is neither of them and the function is neither.
"Neither" is a real answer and students avoid giving it, because the question offered two names and they feel obliged to pick one. Most functions are neither. The names describe a symmetry, and symmetry is the exception.
Odd does not mean odd powers, and even does not mean even powers. It usually works out that way for polynomials, which is why the names were chosen, and it fails as soon as the function is not a polynomial: cos x is even and sin x is odd, and neither has a power in it anywhere.
It also fails for a polynomial with mixed powers. x² + x has one even power and one odd, and the function is neither, because the test is about the whole function and not its terms. Do the test.
A self-inverse function is its own inverse: f(f(x)) = x for every x in the domain. Doing it twice gets you back where you started.
The two the course expects you to recognise are 1/x, which you met at 2.8, and
f(x) = (x + 1)/(x − 1)
Check it: f(4) = 5/3 = 1.667, and f(1.667) = 4. Or more cleanly, f(3) = 2 and f(2) = 3: those two simply swap.
What it looks like on a graph. An inverse is a reflection in y = x, so a self-inverse function is unchanged by that reflection: its graph is symmetric about y = x. That is the fastest way to recognise one.
And it shows in the asymptotes. Reflecting in y = x swaps the vertical and horizontal asymptotes, so a self-inverse rational function must have them matching. Here the vertical is x = 1 and the horizontal is y = 1, the same number arrived at two independent ways: one from the zero of the denominator, one from the behaviour at infinity.
Every claim on this page is a two-line numerical check, which is exactly what a calculator is for. It will not prove a symmetry, because checking values is not a proof, but it will catch a wrong answer.
When you may use it. Analysis Paper 1 is non-calculator, and the algebra of f(−x) is Paper 1 work. Use the machine while learning, to build the habit of testing, and then do the algebra.
The mark people lose. Finding the inverse of a quadratic and not stating the restricted domain. The algebra is usually right and the answer is incomplete, which is the most annoying way to lose a mark. The habit: whenever you take a square root while inverting, write down which half you kept and why. One line, one mark.
1. Find √((−3)²).
2. For f(x) = x³, find f(−4).
3. For g(x) = x² + x, find g(−2).
4. For h(x) = (x + 1)/(x − 1), find h(3).
5. Why does x² have no inverse unless its domain is restricted?
1 markThe test stated, f(−x) compared with f(x) or −f(x).
1 markThe algebra of f(−x), simplified correctly.
1 markThe conclusion, naming odd, even or neither.
On an inverse question add a mark for the restricted domain, stated as part of the answer and not as an afterthought. "f⁻¹(x) = √x for x ≥ 0" is the full answer; the formula alone is not.
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