Equal rational functions: the same domain and the same rule in simplest form
Mathematics Prep 3 Term 1, Unit 2, Lesson 4
GoalDecide whether two rational functions are equal by comparing their domains and their rules in simplest form.
New words
equal functions f = g · functions with the same domain and the same value at every x of that domainsimplest form · the rule after cancelling every common factorsame domain · the sets Df and Dg are exactly equalcounterexample · one value of x that shows two functions are different
Explanation
Two functions are equal, f = g, when two things hold: (1) they have the same domain, Df = Dg, and (2) their rules, written in simplest form, are identical. To test, find each domain from the ORIGINAL denominator, then simplify each rule and compare. If the simplified rules agree but the domains differ, the functions are not equal, even though they give the same value wherever both are defined. A single value of x in one domain but not in the other is enough to show that they are different, as with f(x) = (x² − 4) ÷ (x − 2) and g(x) = x + 2: f(2) does not exist but g(2) = 4.
Condition
How to check it
1. same domain
zeros of the original denominators
2. same rule
factorise, cancel, compare
both true
f = g
one fails
f ≠ g
Worked examples
Are f(x) = (x² − 2x) ÷ (x² − 4x + 4) and g(x) = 3x ÷ (3x − 6) equal? For f: x² − 4x + 4 = (x − 2)², so Df = R − {2}, and f(x) = x(x − 2) ÷ (x − 2)² = x ÷ (x − 2). For g: 3x − 6 = 0 at x = 2, so Dg = R − {2}, and g(x) = 3x ÷ (3(x − 2)) = x ÷ (x − 2). Same domain, same rule: f = g.
Are f(x) = (x² − 4) ÷ (x − 2) and g(x) = x + 2 equal? f simplifies to x + 2, the same rule as g. But Df = R − {2} and Dg = R. The domains differ, so f ≠ g. Proof: g(2) = 4 exists, f(2) does not.
Is f(x) = 1 equal to g(x) = (x² + 1) ÷ (x² + 1)? x² + 1 is never 0, so Dg = R = Df, and g(x) = 1. They are equal. But h(x) = x ÷ x is different from f: Dh = R − {0}.
Class activity
Equal or not? Make three pairs of functions of your own: one pair that is equal, one pair with the same rule but different domains, and one pair with different rules. Swap with a partner to sort them.
Practice
1
Are (x² − 1) ÷ (x − 1) and x + 1 equal?
No. The first has domain R − {1} and the second has domain R.
2
Are x²÷x² and 1 equal?
No. x² ÷ x² has domain R − {0}, while 1 has domain R.
3
Are (x² + 4) ÷ (x² + 4) and 1 equal?
Yes. x² + 4 is never 0, so both have domain R and the same rule 1.
4
Are (x − 3) ÷ (3 − x) and −1 equal?
No. The first simplifies to −1 but its domain is R − {3}; the domain of −1 is R.
5
Are (x² − 3x) ÷ (x² − 6x + 9) and 4x ÷ (4x − 12) equal?
Both have domain R − {3} and both simplify to x ÷ (x − 3). Yes.
