Functions and Graphs
What the domain tests
Functions and graphs examines function notation, domain and range, composite and inverse functions, and graph transformations. MSCE Additional Mathematics items often require you to explain why an inverse exists or to deduce a graph’s behaviour from an equation rather than from a plotted curve.
Core rules
- The composite fg(x) means f(g(x)); the inner function is applied first. Its domain is the set of x in the domain of g for which g(x) lies in the domain of f.
- An inverse f^{-1} exists only when f is one-to-one on its stated domain. To find it, write y=f(x), swap x and y, solve for y, then state the domain of f^{-1}, which is the range of f.
- Transformations: y=f(x)+a shifts up by a; y=f(x-a) shifts right by a; y=af(x) stretches vertically by factor a; y=f(ax) stretches horizontally by factor 1/a; y=-f(x) reflects in the x-axis; y=f(-x) reflects in the y-axis.
- For y=ax^2+bx+c, completing the square locates the vertex; the axis of symmetry is x=-b/(2a). For y=a/(x-h)+k, the asymptotes are x=h and y=k.
Worked example
Let f(x)=2x-5 for x≥0. Find f^{-1} and its domain. Set y=2x-5. Swap x and y: x=2y-5. Solving gives y=(x+5)/2. Because x≥0, f(x)≥-5, so the range of f is y≥-5. Hence f^{-1}(x)=(x+5)/2 with domain x≥-5.
Traps
Omitting the restricted domain; reversing the order in a composite; reflecting in the wrong axis when the sign is inside the bracket; quoting the range of f as the domain of f^{-1} without checking endpoints.