Algebra and Functions
Core ideas
A function assigns exactly one output to each input; the domain is the set of permitted inputs and the range the resulting outputs. Candidates must handle notation such as f(x), composite functions fg(x) meaning f(g(x)), and inverse functions f⁻¹. For an inverse to exist the function must be one-to-one. To find an inverse, write y = f(x), rearrange to make x the subject, then swap x and y. Also required: completing the square, the discriminant, factor and remainder theorems, and sketching quadratics showing intercepts and turning point.
Traps
- Confusing fg(x) with gf(x); order matters.
- Forgetting to restrict the domain of f⁻¹ to the range of f.
- Sign errors when completing the square: x² + bx = (x + b/2)² − b²/4.
- Assuming every quadratic has real roots; check the discriminant b² − 4ac.
- Losing the negative root when solving by square-rooting.
Worked example
Find the inverse of f(x) = (2x + 3)/(x − 1), x ≠ 1. Set y = (2x+3)/(x−1). Then y(x−1) = 2x+3, so yx − y = 2x + 3, giving x(y − 2) = y + 3, hence x = (y+3)/(y−2). Therefore f⁻¹(x) = (x+3)/(x−2), x ≠ 2. The excluded value transfers: the range of f becomes the domain of f⁻¹.
Rule of thumb
Whenever you form an inverse, immediately state its domain by quoting the range of the original function.