3
Zeros of a rational function and the solution set of f(x) = k
Mathematics Prep 3 Term 1, Unit 2, Lesson 3
GoalFind the zeros of a rational function and solve f(x) = k, rejecting any value outside the domain.
New words
zero of f · a value x in the domain with f(x) = 0set of zeros · zeros of the numerator, without the zeros of the denominatorsolution set · all values of x in the domain that make the equation truereject · throw away a value that is outside the domain
Explanation
A fraction is 0 only when its numerator is 0, and its denominator is not. So the set of zeros of f = p ÷ q is {zeros of p} − {zeros of q}. A value that makes both top and bottom 0 is not a zero, because f is undefined there. To solve f(x) = k, first find the domain, then simplify, then solve the simpler equation. At the end check each answer against the domain: an answer outside the domain is rejected, and if none is left the solution set is ∅. When the fraction is not simplified, you may multiply both sides by the denominator, since it is not zero inside the domain.
| Step | For f(x) = (x² − 6x + 8) ÷ (x² − x − 2) |
|---|---|
| zeros of numerator | (x − 2)(x − 4): {2, 4} |
| zeros of denominator | (x − 2)(x + 1): {2, −1} |
| domain | R − {2, −1} |
| zeros of f | {2, 4} − {2, −1} = {4} |
Worked examples
Find the zeros of f(x) = (x² − 6x + 8) ÷ (x² − x − 2). Numerator (x − 2)(x − 4) has zeros {2, 4}. Denominator (x − 2)(x + 1) has zeros {2, −1}. At 2 both are 0, so 2 is not a zero of f. The set of zeros is {4}. Check: f(4) = (16 − 24 + 8) ÷ (16 − 4 − 2) = 0 ÷ 10 = 0.
Let f(x) = (x² − x − 6) ÷ (x − 3). Solve f(x) = 5 and f(x) = 4. Domain R − {3}. Simplify: (x − 3)(x + 2) ÷ (x − 3) = x + 2. For f(x) = 5: x + 2 = 5 gives x = 3, which is not in the domain, so the solution set is ∅. For f(x) = 4: x + 2 = 4 gives x = 2, in the domain. Check: f(2) = (4 − 2 − 6) ÷ (−1) = 4 ✓. Solution set {2}.
Solve (2x + 3) ÷ (x − 1) = 1. Domain R − {1}. Multiply both sides by x − 1: 2x + 3 = x − 1, so x = −4. It is in the domain. Check: (−8 + 3) ÷ (−4 − 1) = −5 ÷ −5 = 1 ✓.
Class activity
Why reject? Take f(x) = (x² − 9) ÷ (x − 3). Try to solve f(x) = 6. Show that the working leads to x = 3 and explain, in your own words, why 3 cannot be an answer.
Practice
1
Find the zeros of (x² − 9) ÷ (x + 3).
Numerator zeros {3, −3}, denominator zeros {−3}. Set of zeros {3}.
2
Find the zeros and the domain of (x² + x) ÷ (x² + 1).
x(x + 1) = 0 gives {0, −1}. x² + 1 is never 0, so the domain is R and the zeros are {0, −1}.
3
Solve (x² − 4) ÷ (x − 2) = 4.
Simplify to x + 2 = 4, so x = 2, but 2 is not in the domain R − {2}. The solution set is ∅.
4
Solve (x² − 4) ÷ (x − 2) = 1.
x + 2 = 1 gives x = −1, which is in the domain. Check: (1 − 4) ÷ (−3) = 1. Solution set {−1}.
5
Solve 3 ÷ (x + 1) = 1/2.
Domain R − {−1}. Cross-multiply: 6 = x + 1, so x = 5. Check: 3 ÷ 6 = 1/2 ✓.