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Review: rational algebraic functions, their domains, simplifying and operations
Mathematics Prep 3 Term 1, Units 1–4, Review
GoalMix the ideas of rational functions: domain, simplifying, holes and operations.
New words
domain · the values of x allowed: where the denominator is not 0simplify · cancel common factors (after factorising)hole · a missing point of the graph at a cancelled factordivide fractions · multiply by the reciprocal, and also keep the divisor’s numerator ≠ 0
Explanation
A rational function is a fraction of polynomials. Find the domain first, from the ORIGINAL denominator, and only then cancel. Example f(x) = (x² − 9) ÷ (x² − 5x + 6) = (x − 3)(x + 3) ÷ ((x − 2)(x − 3)). The domain is R − {2, 3}; the dashed line x = 2 marks where the graph has no point at all. Cancelling gives (x + 3) ÷ (x − 2), but x = 3 stays excluded: at x = 3 the graph has a hole at (3, 6). Adding or subtracting needs a common denominator, and the sum x/(x − 3) − 3/(x − 3) = (x − 3)/(x − 3) = 1 is only true for x ≠ 3. Dividing: multiply by the reciprocal, and the domain also excludes where the divisor is 0.
Worked examples
Simplify (x² − x − 6) ÷ (x² − 9) = (x − 3)(x + 2) ÷ ((x − 3)(x + 3)) = (x + 2) ÷ (x + 3), for x ≠ 3 and x ≠ −3.
Divide ((x² − 4) ÷ (x + 3)) ÷ ((x + 2) ÷ (x² − 9)) = (x − 2)(x + 2)/(x + 3) × (x − 3)(x + 3)/(x + 2) = (x − 2)(x − 3). Domain: x ≠ −3, −2, 3.
Subtract x/(x − 3) − 3/(x − 3) = (x − 3)/(x − 3) = 1, domain R − {3}.
Class activity
Hole hunter. Choose a function such as (x − 1)(x + 2) ÷ (x − 1). Find where the hole is and what its height is.
Practice
1
Find the domain of f(x) = (x + 1) ÷ (x² − 16).
x² − 16 ≠ 0, so R − {−4, 4}.
2
Simplify (x² − 4) ÷ (x − 2) and give the hole.
x + 2 for x ≠ 2; hole at (2, 4).
3
Add 1/(x − 1) + 2/(x − 1).
3/(x − 1), for x ≠ 1.
4
Are f(x) = (x² − 1)/(x − 1) and g(x) = x + 1 equal? Why?
No: their domains differ (f excludes 1).
5
Multiply (x/(x + 1)) × ((x + 1)/(x − 2)).
x/(x − 2), with x ≠ −1, 2.