Algebra and Functions
What this domain tests
Real functions: domain and range, injectivity, inverse, composition, monotonicity, graph transformations, and polynomial, rational, exponential, logarithmic and absolute-value equations and inequalities. Marks are given for exact set notation and sign tables, not for sketches.
Core rules
- Domain: exclude zeros of denominators, require even-root radicands ≥ 0 and log arguments > 0. Find the range from monotonicity or from the inverse.
- f is injective when f(x₁) = f(x₂) forces x₁ = x₂; strictly monotone on an interval implies injective there. Only injective functions have inverses, and the graphs of f and f⁻¹ are mirror images in y = x, with domain and range interchanged.
- For (g∘f)(x) = g(f(x)) you need x ∈ D_f and f(x) ∈ D_g; the second condition is where marks are lost.
- Quadratic ax² + bx + c: if a > 0 and Δ < 0 the expression is positive for all x; otherwise use the sign table - same sign as a outside the roots, opposite between them.
- a^x and log_a x are inverse functions; a > 1 gives increasing, 0 < a < 1 decreasing. |x| < k means −k < x < k, while |x| > k splits into two branches.
Worked example
Find the domain and inverse of f(x) = ln((1+x)/(1−x)).
The logarithm needs (1+x)/(1−x) > 0. The numerator vanishes at x = −1, the denominator at x = 1; the sign table gives positivity exactly on (−1, 1), so D_f = (−1, 1). There the fraction is strictly increasing and ln is increasing, so f is injective; as x → −1⁺ the argument → 0⁺ and f → −∞, as x → 1⁻ it → +∞, so the range is ℝ.
Set y = ln((1+x)/(1−x)). Then e^y = (1+x)/(1−x), so e^y − e^y·x = 1 + x, giving e^y − 1 = x(1 + e^y), hence x = (e^y − 1)/(e^y + 1). Therefore f⁻¹(x) = (e^x − 1)/(e^x + 1), with domain ℝ and range (−1, 1) - the swapped pair, as required.
Traps that cost marks
- Writing the domain of a composition without checking that f(x) lands inside D_g.
- Squaring both sides and keeping spurious roots; always substitute candidates back.
- Multiplying an inequality by (x + 1) or (1 − x) without splitting into cases by its sign.
- Claiming f(x) = x² has an inverse on ℝ; restrict to [0, +∞) first.
- Confusing the range of ln (all of ℝ) with its domain (positive reals).