Can restricting the domain change the parity of an unchanged formula?
My prediction and reason
In the projection model, predict whether x² is even on the reals and on the restricted domain [0, 3]. Then predict whether the zero function on the reals is odd, even, both or neither.
Evidence and observations
Choose x² with the symmetric real domain in the projection model. Record paired inputs ±1 and ±2 and their outputs, then compute f(−x) for arbitrary x.
Switch the x² projection model from the real domain to [0, 3]. Test x = 1 and its reflected input −1; record which is excluded after the switch.
For 1/x and the zero function, write both domain-symmetry checks and the equations for f(−x). Record whether zero must be in the domain.
Explain or give a counterexample
Transfer
Classify x³ + x on the reals, x² + x on the reals, and x² on [1, 3]. For each, state the domain check before the algebraic test.
Exit task
Transfer to a different restricted domain: is f(x) = x² even on [0, 2]? Is z(x) = 0 on the reals both odd and even? Give a reason for each.
5 min predict domain and zero-function cases; 12 min compare reflected pairs; 8 min establish identities; 12 min paper classification; 3 min exit check.
Can restricting the domain change the parity of an unchanged formula?
In the projection model, predict whether x² is even on the reals and on the restricted domain [0, 3]. Then predict whether the zero function on the reals is odd, even, both or neither.
Evidence tasks
- Choose x² with the symmetric real domain in the projection model. Record paired inputs ±1 and ±2 and their outputs, then compute f(−x) for arbitrary x.
- Switch the x² projection model from the real domain to [0, 3]. Test x = 1 and its reflected input −1; record which is excluded after the switch.
- For 1/x and the zero function, write both domain-symmetry checks and the equations for f(−x). Record whether zero must be in the domain.
Reference reasoning
Parity requires a domain symmetric about zero. On such a domain, even means f(−x) = f(x) for every input and gives reflection in the vertical axis; odd means f(−x) = −f(x) and gives symmetry about the origin. In the projection model, x² is even on R but not on [0, 3], because −1 is missing when 1 is allowed. The paper’s x² + 1 example is another even function on R, proved by its own identity. The punctured domain of 1/x is symmetric, so it can be odd without being defined at zero. The zero function satisfies both identities. On a symmetric domain, a function that is both odd and even has f(x) = −f(x), hence f(x) = 0 everywhere.
Paper example: classify x² + 1 on the reals; the projection’s square rule has no added constant.
f(x)=x2+1,x∈R - The domain R is symmetric about zero.
- f(−x) = (−x)² + 1 = x² + 1 = f(x) for every real x. It is even, and is not odd since f(0) = 1.
Classify the reciprocal function, including its domain condition.
f(x)=x1,x∈R∖{0} - If x ≠ 0, then −x ≠ 0, so the domain is symmetric.
- f(−x) = 1/(−x) = −1/x = −f(x). It is odd; it need not be defined at zero.
Transfer and exit task
Classify x³ + x on the reals, x² + x on the reals, and x² on [1, 3]. For each, state the domain check before the algebraic test.
Transfer to a different restricted domain: is f(x) = x² even on [0, 2]? Is z(x) = 0 on the reals both odd and even? Give a reason for each.
f is not even because its domain is not symmetric: 1 is allowed but −1 is not. z is both, since z(−x) = 0 = z(x) = −z(x) and its domain is symmetric.
- Checks the domain rather than relying only on the square formula.
- Explicitly verifies both zero-function identities.
Questions
- After restricting the projection domain to [0, 3], what happens to the point with input −1?
- Where would an odd function defined at zero place its point above zero?
Expected responses
- Students may decide parity from the exponent alone. Compare the same formula with two domains.
- Some think odd and even are mutually exclusive. Substitute the zero function into both definitions.
Confusions
- Paired numerical samples support a conjecture; arbitrary-x identities establish parity.
- Odd symmetry is about the origin, not a reflection in the horizontal axis alone.
Board plan
- First: domain symmetric about zero.
- Even: f(−x) = f(x); odd: f(−x) = −f(x), for all allowed x.
- Zero function on a symmetric domain: both odd and even.
