Week 5 · Common domain and equivalent rational algebraic functions · 3 lessons
3
Equal only on the common domain: proving f(x) = g(x) there, and showing f ≠ g
Mathematics Prep 3 Term 1, Unit 2, Lesson 4
GoalProve that two functions agree on their common domain, show that they are still different functions, and solve f(x) = g(x).
New words
common domain · where both functions are definedprove · show with steps that a statement is always trueagree · give the same value at the same xreject · drop an answer that lies outside the common domain
Explanation
Two functions with the same simplest rule but different domains are not equal, but they do agree on the part of the domain they share. To prove f(x) = g(x) on the common domain: (1) find Df, Dg and the common domain; (2) simplify both rules; (3) conclude they are the same rule, so the two functions give the same value at every x they share. To show f ≠ g, point to a value of x in one domain only. To solve f(x) = g(x) when the rules are different, work only in the common domain, solve the equation, and reject any answer that is not in it. If none is left, the solution set is ∅.
x
−1
0
1
2
3
f(x)
−3/2
−3
no value
3
3/2
g(x)
−3/2
no value
no value
3
3/2
Worked examples
Let f(x) = 3 ÷ (x − 1) and g(x) = 3x ÷ (x² − x). Prove f(x) = g(x) on the common domain and show f ≠ g. Df = R − {1}. x² − x = x(x − 1), so Dg = R − {0, 1}. Common domain R − {0, 1}. g(x) = 3x ÷ (x(x − 1)) = 3 ÷ (x − 1) = f(x) there. But 0 is in Df and not in Dg: f(0) = −3 exists, g(0) does not. So f ≠ g.
Solve f(x) = g(x) for f(x) = (x + 4) ÷ (x − 2) and g(x) = (2x + 3) ÷ (x − 2). The common domain is R − {2}. Same denominator, so x + 4 = 2x + 3 and x = 1. It is in the domain. Check: f(1) = 5 ÷ (−1) = −5 and g(1) = 5 ÷ (−1) = −5 ✓. Solution set {1}.
Look at the table: f(x) = 3 ÷ (x − 1) and g(x) = 3x ÷ (x² − x). At x = 0 only f has a value, at x = 1 neither has one, and everywhere else the two rows agree. That is exactly the statement: equal on the common domain, but not the same function.
Class activity
Spot the mistake. A student says: “f(x) = x ÷ (x² − 2x) and g(x) = 1 ÷ (x − 2) have the same simplest rule, so f = g.” Find the domains of both and explain what the student forgot. On which set do they agree?
Practice
1
Show that f and g agree wherever both are defined, for f(x) = (x² − 1) ÷ (x² − x) and g(x) = (x + 1) ÷ x. Then say whether f = g.
Df = R − {0, 1} and Dg = R − {0}. On R − {0, 1}: f(x) = (x − 1)(x + 1) ÷ (x(x − 1)) = (x + 1) ÷ x = g(x). But g(1) = 2 exists and f(1) does not, so f ≠ g.
2
Solve f(x) = g(x) for f(x) = x ÷ (x − 3) and g(x) = 3 ÷ (x − 3).
Common domain R − {3}. x = 3, which is not in the domain. The solution set is ∅.
3
Solve f(x) = g(x) for f(x) = 2x ÷ (x + 1) and g(x) = 6 ÷ (x + 1).
Common domain R − {−1}. 2x = 6, so x = 3, in the domain. Check: 6 ÷ 4 = 6 ÷ 4 ✓. Solution set {3}.
4
At which x is f(x) = (x² − 4) ÷ (x − 2) different from g(x) = x + 2?
Only at x = 2: f(2) is undefined and g(2) = 4. Everywhere else they agree.
5
Are f(x) = 5 and g(x) = 5x ÷ x equal on the common domain? Are they equal functions?
Common domain R − {0}, and g(x) = 5 there, so they agree on it. They are not equal functions: Df = R but Dg = R − {0}.
