Functions and graphs
What this domain covers
Function questions in 8227 test whether you can state a domain and range precisely, decide whether a function is one-one, form and evaluate composite functions, find an inverse together with its domain, and describe transformations of standard curves.
Core rules
- Domain: exclude values that make a denominator zero, make an even root's argument negative, or make a logarithm's argument non-positive. The range follows from the graph on that restricted domain, and completing the square is often the fastest route.
- fg(x) means f(g(x)): apply g first. It exists only for x in the domain of g whose image g(x) lies in the domain of f.
- f⁻¹ exists only when f is one-one. Write y = f(x), rearrange to x in terms of y, then swap x and y. Domain of f⁻¹ = range of f, and the two graphs are reflections in y = x.
- Transformations: f(x) + a shifts up by a; f(x + a) shifts left by a; af(x) stretches vertically by factor a; f(ax) stretches horizontally by factor 1/a; −f(x) and f(−x) reflect in the x-axis and y-axis.
Worked example
f(x) = x² − 4x + 7 for x ≥ 2. Completing the square gives f(x) = (x − 2)² + 3. On x ≥ 2 the least value is 3, attained at x = 2, so the range is f(x) ≥ 3 and f is one-one. For the inverse, y − 3 = (x − 2)², so x − 2 = ±√(y − 3); the restriction x ≥ 2 forces the positive sign. Hence f⁻¹(x) = 2 + √(x − 3) with domain x ≥ 3.
Traps
- Writing f⁻¹(x) = 1/f(x).
- Ignoring the stated domain, then losing or duplicating the ± choice in the inverse.
- Applying a horizontal stretch in the wrong order relative to a translation.
- Quoting a range as an inequality without confirming the endpoint is actually attained.