Polynomials
Senior 2 · Mathematics
Polynomials
Lesson 3 of 5
22 min
A function is a rule that turns each input into exactly one output. When the rule is a polynomial, it is a polynomial function. Real processes often happen in stages: a shop first converts a price in dollars to francs, then adds a delivery charge. Doing one function after another is called composition.
Sometimes you need to run a rule backwards: if a moto fare was 1,500 Frw, how far was the trip? Running a function backwards means finding its inverse.
f(x) = 2x + 1 is a linear polynomial function and g(x) = x2 − 3 is a quadratic one. f(4) = 2(4) + 1 = 9 is the numerical value of f at 4.
The composite function f ∘ g (read "f of g" or "f after g") is defined by
(f ∘ g)(x) = f(g(x)): first apply g, then apply f to the result.
Worked example 1
Let f(x) = 2x + 1 and g(x) = x2 − 3. Find (f ∘ g)(x), (g ∘ f)(x) and (f ∘ g)(2).
(f ∘ g)(x) = f(x2 − 3) = 2(x2 − 3) + 1 = 2x2 − 5.
(g ∘ f)(x) = g(2x + 1) = (2x + 1)2 − 3 = 4x2 + 4x + 1 − 3 = 4x2 + 4x − 2.
(f ∘ g)(2) = 2(4) − 5 = 3. Check step by step: g(2) = 4 − 3 = 1, then f(1) = 3 ✓.
The two composites are different polynomials, so the order matters: in general f ∘ g ≠ g ∘ f.
The inverse f−1 undoes f: if f(a) = b, then f−1(b) = a. Every linear function f(x) = ax + b with a ≠ 0 has an inverse, found in three steps.
Worked example 2
Find the inverse of f(x) = 3x − 4.
Check: f(3) = 9 − 4 = 5, and f−1(5) = 9 ÷ 3 = 3 ✓. The inverse brought 5 back to 3.
Worked example 3
(a Rwandan context). A moto company charges a fixed 300 Frw plus 100 Frw per kilometre, so the fare is F(d) = 100d + 300. Find F−1 and use it for a fare of 1,500 Frw.
y = 100d + 300 gives d = (y − 300) ÷ 100, so F−1(x) = (x − 300) ÷ 100. Then F−1(1500) = 1200 ÷ 100 = 12. The trip was 12 km.
Tip
Memory hook. Read f ∘ g from right to left: the function nearest to x acts first, like putting on socks (g) before shoes (f). For an inverse, undo the steps in reverse order: "×3 then −4" is undone by "+4 then ÷3".
f(x) = x + 5 and g(x) = 3x. Find (f ∘ g)(x) and (g ∘ f)(x).
Solution: (f ∘ g)(x) = 3x + 5. (g ∘ f)(x) = 3(x + 5) = 3x + 15.
With f(x) = 2x + 1 and g(x) = x2 − 3, find (g ∘ f)(2).
Solution: f(2) = 5, then g(5) = 25 − 3 = 22. (Also from the formula: 4(4) + 4(2) − 2 = 22.)
Find the inverse of h(x) = 5x + 2.
Solution: y = 5x + 2 gives x = (y − 2) ÷ 5, so h−1(x) = (x − 2) ÷ 5.
If f(x) = 4x − 7, find f−1(9).
Solution: Solve 4x − 7 = 9: 4x = 16, x = 4. So f−1(9) = 4.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
