1.2 Functions
- Syllabus
- 9709–2028–2029
- Topic
- 1.2
- Level
- AS
A function f maps each permitted input x to exactly one output f(x). The domain is the allowed set of inputs; the range is the set of outputs actually produced.
Read f(a) as the output when input a is used. For composites, (f∘g)(x)=f(g(x)) and x must lie in g’s domain with g(x) in f’s domain.
For f(x)=√(x−2), the real domain is x≥2 and the range is y≥0; f(6)=2.
A relation can assign one input several outputs and then is not a function; domain restrictions are part of the definition, not optional notes.
The range is the set of outputs produced by a function on its stated domain. For (f∘g)(x)=f(g(x)), x must be allowed by g and g(x) must lie in f’s domain.
Find the inner range first, then apply the outer function. Restricting a domain can change the range and can make a formula that is usually invertible behave differently.
If g(x)=x² on [0,2] and f(u)=√u, then (f∘g)(x)=x on [0,2]; using all real x would describe a different domain and range.
The range of f∘g is not automatically the range of f; only the part of f reached by g is relevant.
A function is one-one if f(a)=f(b) implies a=b. Equivalently, no horizontal line meets its graph more than once. This allows an inverse function to undo f on its range.
To find f⁻¹, write y=f(x), interchange x and y, then solve for y. The inverse domain is the original range, and its range is the original domain.
f(x)=3x−2 is one-one on ℝ and f⁻¹(x)=(x+2)/3. The graphs reflect in y=x.
A function can have an inverse relation but not an inverse function if it is not one-one on the stated domain.
If (a,b) lies on y=f(x), then (b,a) lies on y=f−1(x). Swapping coordinates reflects the entire graph in the mirror line y=x.
Draw y=x, reflect several defining points and preserve their order. An x-intercept (a,0) becomes the y-intercept (0,a); horizontal and vertical asymptotes swap; the domain and range swap.
The two graphs can meet only on y=x, because a reflected point is unchanged there. Such intersections satisfy f(x)=x, provided x lies in both relevant domains.
Use the original graph's endpoints, open/closed points, turning behaviour and asymptotes rather than inventing a new shape. The inverse must still pass the vertical-line test because the original is one-one.
Do not reflect in the x-axis or y-axis. A correct inverse sketch must show the relevant mirror line y=x and swap coordinates, not merely reverse the curve's direction.
| New graph | Coordinate mapping from (x,y) | Description |
|---|---|---|
| y=f(x)+a | (x,y+a) | translate by (0,a) |
| y=f(x+a) | (x−a,y) | translate by (−a,0) |
| y=af(x) | (x,ay) | stretch parallel to y-axis, factor ∣a∣; reflect in x-axis if a<0 |
| y=f(ax) | (x/a,y) | stretch parallel to x-axis, factor 1/∣a∣; reflect in y-axis if a<0 |
Track a known point, intercept, asymptote or turning point through the coordinate mapping. This is safer than relying on a memorised verbal direction, especially for inside transformations.
For $y=3f(-2x)+4$: first map $x\mapsto -x/2$ (horizontal reflection and factor $1/2$), then $y\mapsto3y+4$ (vertical factor $3$, then translate up $4$).
When two transformations share one axis, order can change the translation size. State a valid sequence explicitly and verify it by applying the operations to the function or to a reference point.
Inside signs and factors act inversely: f(x−3) moves right 3, while f(3x) has horizontal scale factor 1/3. Use the syllabus terms translation, reflection and stretch with axis/direction and factor.