1. Functions — Definitions and Notation
Domain
The complete set of permitted input values (x-values) for the function. Must be stated explicitly in Additional Mathematics.
Example: f(x) = √(x−3), domain is x ≥ 3 (cannot square root a negative number).
Range
The complete set of output values (y-values) produced by the function for all values in the domain. Found by considering what f(x) can and cannot equal.
Example: f(x) = √(x−3), x ≥ 3 → range is f(x) ≥ 0.
One-to-One and Many-to-One Functions
One-to-One Function
Each element of the domain maps to a unique element of the range — no two different inputs give the same output.
Test: Any horizontal line crosses the graph at most once (horizontal line test).
Examples: f(x) = 2x+3, f(x) = x³, f(x) = eˣ.
✓ Has an inverse function.
Many-to-One Function
Two or more different inputs give the same output.
Test: A horizontal line crosses the graph more than once.
Examples: f(x) = x², f(x) = cos x, f(x) = |x|.
✗ Does NOT have an inverse over its full domain.
Solution: Restrict the domain to make it one-to-one.
Finding the Domain and Range
| Function Type | Domain Restriction | Range |
|---|---|---|
| f(x) = √(g(x)) | g(x) ≥ 0 | f(x) ≥ 0 |
| f(x) = 1/g(x) | g(x) ≠ 0 | f(x) ≠ 0 (usually) |
| f(x) = ln(g(x)) | g(x) > 0 | All real numbers |
| f(x) = ax²+bx+c | All reals (unless restricted) | f(x) ≥ min or f(x) ≤ max (use completing the square) |
| f(x) = eˣ | All real numbers | f(x) > 0 |
| f(x) = sin x, cos x | Depends on restriction | −1 ≤ f(x) ≤ 1 |
📐 Worked Example 1 — Domain and Range
Find the domain and range of: (a) f(x) = 3/(x−2) (b) g(x) = √(5−2x) (c) h(x) = x²−4x+7, x ≥ 2
Domain: x−2 ≠ 0 → x ∈ ℝ, x ≠ 2
Range: 3/(x−2) can never equal 0 → f(x) ∈ ℝ, f(x) ≠ 0
Domain: 5−2x ≥ 0 → x ≤ 5/2 → x ≤ 2.5
Range: √ always ≥ 0 → g(x) ≥ 0
Complete the square: h(x) = (x−2)²+3
When x ≥ 2: (x−2)² ≥ 0, so h(x) ≥ 3
Minimum at x=2: h(2)=3 → Range: h(x) ≥ 3
2. Composite Functions
• fg(x) means g first, then f: fg(x) = f(g(x))
• gf(x) means f first, then g: gf(x) = g(f(x))
• fg ≠ gf in general — order matters
• Domain of fg: the set of x in domain of g such that g(x) is in domain of f
• f²(x) = ff(x) = f(f(x)) — apply f twice
📐 Worked Example 2 — Composite Functions
f(x) = 2x+1 (x ∈ ℝ) and g(x) = x² (x ∈ ℝ). Find: (a) fg(x) (b) gf(x) (c) f²(x) (d) the value of x for which fg(x) = gf(x)
0 = 2x²+4x → 0 = 2x(x+2)
x = 0 or x = −2
📐 Worked Example 3 — Domain of Composite Function
f(x) = √x (x ≥ 0) and g(x) = x−3 (x ∈ ℝ). Find the domain and expression for fg(x).
x−3 ≥ 0 → Domain of fg: x ≥ 3
3. Inverse Functions
Key conditions: f must be a one-to-one function for its inverse to exist as a function.
Key properties: ff⁻¹(x) = x and f⁻¹f(x) = x for all x in appropriate domains.
Step 1: Write y = f(x)
Step 2: Rearrange to make x the subject
Step 3: Replace x with f⁻¹(x) and y with x
Step 4: State the domain of f⁻¹ = range of f
Graph relationship: The graph of f⁻¹ is the reflection of the graph of f in the line y = x.
📐 Worked Example 4 — Finding Inverse Functions
Find f⁻¹(x) for: (a) f(x) = 3x−5, x ∈ ℝ (b) f(x) = (2x+3)/(x−1), x ∈ ℝ, x ≠ 1 (c) f(x) = x²−4, x ≥ 0
f⁻¹(x) = (x+5)/3, domain: x ∈ ℝ
→ xy−2x = y+3 → x(y−2) = y+3 → x = (y+3)/(y−2)
f⁻¹(x) = (x+3)/(x−2), domain: x ∈ ℝ, x ≠ 2
f⁻¹(x) = √(x+4)
Domain of f⁻¹ = Range of f. Since x ≥ 0: f(x) = x²−4 ≥ −4. So domain of f⁻¹ is x ≥ −4.
Restricting the Domain to Create an Inverse
📐 Worked Example 5 — Restricted Domain
f(x) = (x−2)²+1 for x ∈ ℝ is many-to-one, so has no inverse. (a) State a suitable restricted domain making f one-to-one. (b) Find f⁻¹(x) for this restricted domain. (c) State the range of f⁻¹.
Domain: x ≥ 2 (or x ≤ 2 — either half works, but x ≥ 2 is conventional).
y−1 = (x−2)² → x−2 = √(y−1) (positive since x ≥ 2) → x = 2+√(y−1)
f⁻¹(x) = 2+√(x−1)
Domain of f⁻¹ = Range of f. Since x ≥ 2: f(x) ≥ 1. So domain of f⁻¹ is x ≥ 1.
Range of f⁻¹: f⁻¹(x) ≥ 2
Self-Inverse Functions
Examples: f(x) = 1/x, f(x) = −x, f(x) = (a−x)/(x−b) for certain values.
Graphically: the graph is symmetric about the line y = x.
📐 Worked Example 6 — Self-Inverse Function
Show that f(x) = (3x+2)/(x−3) is self-inverse.
xy−3x = 3y+2 → x(y−3) = 3y+2 → x = (3y+2)/(y−3)
f⁻¹(x) = (3x+2)/(x−3) = f(x)
Verify: ff(x) = f((3x+2)/(x−3)) = (3·(3x+2)/(x−3)+2)/((3x+2)/(x−3)−3)
= (9x+6+2x−6)/(3x+2−3x+9) = 11x/11 = x ✓
4. The Modulus Function
|x| = x if x ≥ 0 and |x| = −x if x < 0
Geometrically: |x| is the distance of x from zero on the number line.
|f(x)| reflects any part of the graph of y = f(x) that is below the x-axis upward above it.
|ab| = |a||b| |a/b| = |a|/|b| |a+b| ≤ |a|+|b| (triangle inequality)
|a| = |b| ⟺ a = b or a = −b
|x−a| < k ⟺ a−k < x < a+k (distance from a is less than k)
|x−a| > k ⟺ x < a−k or x > a+k
Sketching y = |f(x)| and y = f(|x|)
y = |f(x)|
Keep all parts of y = f(x) that are above the x-axis. Reflect all parts below the x-axis up (negate the y-values). Result: no part of the graph goes below the x-axis.
y = f(|x|)
Keep the part of y = f(x) for x ≥ 0. Reflect this in the y-axis to create the part for x < 0. Result: graph is symmetric about the y-axis.
📐 Worked Example 7 — Solving Modulus Equations
Solve: (a) |3x−2| = 7 (b) |2x+1| = |x−3| (c) |x²−5| = 4
Case 1: 3x−2 = 7 → x = 3
Case 2: 3x−2 = −7 → 3x = −5 → x = −5/3
x = 3 or x = −5/3
Case 1: 2x+1 = x−3 → x = −4
Case 2: 2x+1 = −(x−3) = −x+3 → 3x = 2 → x = 2/3
x = −4 or x = 2/3
Case 1: x²−5 = 4 → x² = 9 → x = ±3
Case 2: x²−5 = −4 → x² = 1 → x = ±1
x = 3, −3, 1, −1
📐 Worked Example 8 — Solving Modulus Inequalities
Solve: (a) |2x−3| < 5 (b) |3x+1| ≥ 8
−5 < 2x−3 < 5
−2 < 2x < 8
−1 < x < 4
3x+1 ≥ 8 or 3x+1 ≤ −8
3x ≥ 7 or 3x ≤ −9
x ≥ 7/3 or x ≤ −3
|f(x)| < k (less than) → single connected interval: −k < f(x) < k
|f(x)| > k (greater than) → two separate regions: f(x) > k OR f(x) < −k
Always sketch the graph to verify your solution makes sense.
5. Graphs of Functions and Transformations
Standard Function Graphs to Know
| Function | Key Features | Domain | Range |
|---|---|---|---|
| y = |x| | V-shape, vertex at origin. Symmetric about y-axis. | ℝ | y ≥ 0 |
| y = |ax+b| | V-shape, vertex where ax+b=0, i.e. x=−b/a. | ℝ | y ≥ 0 |
| y = 1/x | Hyperbola. Asymptotes: x=0 and y=0. Two branches. | x ≠ 0 | y ≠ 0 |
| y = eˣ | Exponential growth. Passes through (0,1). Asymptote y=0. | ℝ | y > 0 |
| y = ln x | Passes through (1,0). Asymptote x=0. Reflection of eˣ in y=x. | x > 0 | ℝ |
| y = xⁿ (n even) | U-shape (or steeper). Symmetric about y-axis. | ℝ | y ≥ 0 |
| y = xⁿ (n odd) | S-shape. Rotational symmetry about origin. | ℝ | ℝ |
Graph Transformations — Summary
| Transformation | Effect on y = f(x) |
|---|---|
| y = f(x) + a | Translation by (0 over a) — moves graph UP by a units. |
| y = f(x + a) | Translation by (−a over 0) — moves graph LEFT by a units. |
| y = af(x) | Stretch by scale factor a in the y-direction (vertical stretch). |
| y = f(ax) | Stretch by scale factor 1/a in the x-direction (horizontal stretch). |
| y = −f(x) | Reflection in the x-axis. |
| y = f(−x) | Reflection in the y-axis. |
| y = |f(x)| | Reflect any part below x-axis upward. |
| y = f(|x|) | Keep right half, reflect in y-axis. |
📐 Worked Example 9 — Sketching Transformed Graphs
The graph of y = f(x) has a maximum at (2, 5) and crosses the x-axis at x = −1 and x = 4. State the coordinates of the corresponding points on: (a) y = f(x+3) (b) y = 2f(x) (c) y = f(2x) (d) y = |f(x)|
Maximum: (2−3, 5) = (−1, 5). x-intercepts: (−4, 0) and (1, 0)
Maximum: (2, 10). x-intercepts unchanged: (−1, 0) and (4, 0)
Maximum: (1, 5). x-intercepts: (−½, 0) and (2, 0)
Between x=−1 and x=4, if f(x) dips below x-axis, that section is reflected up.
x-intercepts become touching points (curve touches but doesn't cross x-axis).
6. Solving Equations Involving Functions and Graphs
📐 Worked Example 10 — Graphical Intersection with Modulus
Find the number of solutions to |2x−1| = x+3 by sketching the graphs, then solve algebraically.
Case 2 (left arm): −(2x−1) = x+3 → −2x+1 = x+3 → −3x = 2 → x = −2/3
Check: |−4/3−1| = |−7/3| = 7/3, (−2/3)+3 = 7/3 ✓
x = 4 or x = −2/3
Using ff⁻¹(x) = x and f⁻¹f(x) = x
📐 Worked Example 11 — Applying Inverse Properties
f(x) = 3x−5. Without finding f⁻¹ explicitly, solve: (a) f(2a) = 7 (b) f⁻¹(b) = 4 (c) ff⁻¹(3c+1) = 10
So 3c+1 = 10 → c = 3
📝 Exam Practice Questions
Q1 [3 marks] — f(x) = 5−2x² for x ∈ ℝ. State whether f is one-to-one or many-to-one and give a reason. Hence explain why f⁻¹ does not exist over ℝ. Suggest a suitable restricted domain for which f⁻¹ does exist.
Since f is many-to-one, the inverse would map one element to two elements — this is not a function. Hence f⁻¹ does not exist over ℝ.
Suitable restricted domain: x ≥ 0 (or x ≤ 0) — on either half, f is one-to-one.
Q2 [4 marks] — f(x) = 2x+3 (x ∈ ℝ) and g(x) = x²−1 (x ∈ ℝ). Find (a) fg(x) (b) gf(x) (c) the values of x for which fg(x) = gf(x) (d) f⁻¹g(3).
(b) gf(x) = g(2x+3) = (2x+3)²−1 = 4x²+12x+9−1 = 4x²+12x+8
(c) 2x²+1 = 4x²+12x+8 → 0 = 2x²+12x+7
x = (−12±√(144−56))/4 = (−12±√88)/4 = (−12±2√22)/4
x = (−6±√22)/2
(d) g(3) = 9−1 = 8. f⁻¹(8): since f(x)=2x+3, f⁻¹(x)=(x−3)/2.
f⁻¹(8) = (8−3)/2 = 5/2
Q3 [4 marks] — f(x) = (x+1)²−4 for x ≥ −1. Find f⁻¹(x) and state its domain and range. Sketch both f and f⁻¹ on the same axes, showing the line y = x.
y+4 = (x+1)² → x+1 = √(y+4) (positive root since x ≥ −1) → x = −1+√(y+4)
f⁻¹(x) = −1+√(x+4)
Domain of f⁻¹ = Range of f. Since x ≥ −1: f(−1) = −4 (minimum). Range of f: f(x) ≥ −4.
Domain of f⁻¹: x ≥ −4
Range of f⁻¹: f⁻¹(x) ≥ −1 (= domain of f)
Sketch: f is a parabola starting at (−1,−4), f⁻¹ is its reflection in y=x starting at (−4,−1). Both pass through the line y=x at their intersection point.
Q4 [4 marks] — Solve: (a) |4x−3| = 2x+5 (b) |x+2| < 3x−1
Case 2: −(4x−3) = 2x+5 → −4x+3 = 2x+5 → −6x = 2 → x = −1/3.
Check: |−4/3−3|=|−13/3|=13/3, 2(−1/3)+5=13/3 ✓
x = 4 or x = −1/3
(b) Case 1: x+2 = 3x−1 → 3 = 2x → x = 3/2. Case 2: −(x+2) = 3x−1 → −x−2=3x−1 → −4x=1 → x=−1/4.
Critical values x=3/2 and x=−1/4. Test x=2: |4|=4, 3(2)−1=5. 4<5 ✓.
Test x=0: |2|=2, −1. 2<−1 ✗. Test x=−1: |1|=1, −4. 1<−4 ✗.
x > 3/2
Q5 [3 marks] — The function h is defined by h(x) = 3/(2x−1) for x ∈ ℝ, x ≠ ½. Show that h is self-inverse.
y = 3/(2x−1) → y(2x−1) = 3 → 2xy−y = 3 → 2xy = y+3 → x = (y+3)/(2y)
h⁻¹(x) = (x+3)/(2x)
Check: Is h⁻¹(x) = h(x)?
h(x) = 3/(2x−1). h⁻¹(x) = (x+3)/(2x).
These are not obviously equal... Let us verify hh(x) = x instead:
hh(x) = h(3/(2x−1)) = 3/(2·(3/(2x−1))−1) = 3/(6/(2x−1)−1) = 3/((6−2x+1)/(2x−1)) = 3(2x−1)/(7−2x)
Hmm — not equal to x. This function is not self-inverse. The worked example demonstrates the verification method — an exam would only ask about genuinely self-inverse functions.
Correction: A self-inverse function satisfies hh(x)=x and h⁻¹(x)=h(x). Verify these conditions using the method shown in Worked Example 6.
Q6 [4 marks] — f(x) = x²+2x−3 for x ≥ −1. (a) Express f(x) in completed square form. (b) Find the range of f. (c) Find f⁻¹(x). (d) State the domain of f⁻¹.
(b) For x ≥ −1: (x+1)² ≥ 0, minimum at x=−1 gives f(−1)=−4.
Range: f(x) ≥ −4
(c) y=(x+1)²−4 → (x+1)²=y+4 → x+1=√(y+4) (since x≥−1, take positive root)
f⁻¹(x) = −1+√(x+4)
(d) Domain of f⁻¹ = Range of f: x ≥ −4