Additional Mathematics Synthesis Guide 5: Functions, Inverses, Compositions and Graph Transformations

Input → rule → output → graph → inverse → composition → transformation → verification.

1. Function notation is a map, not decoration

If f(x) = 2x + 3, then f names the function and x is the input placeholder. The statement f(5) = 13 means the machine receives 5 and returns 13. The notation becomes especially powerful when the input is itself an expression: f(a + 1) = 2(a + 1) + 3 = 2a + 5.

One of the most common errors is to treat f(x) like multiplication between f and x. It is not. The parentheses identify the input supplied to the function.

2. Domain and range define the legal operating region

The domain is the set of allowed inputs. The range is the set of outputs actually produced. These are not administrative details. They determine whether an inverse exists as a function and whether an algebraic expression is even meaningful.

• For f(x) = 1/(x − 2), x = 2 is excluded from the domain.
• For g(x) = √(x + 5) over the reals, x ≥ −5.
• For h(x) = x² on all real numbers, the range is h(x) ≥ 0.

A strong student asks domain questions before manipulating aggressively. This prevents later solutions from wandering into values that the original function never allowed.

3. The graph is the visible record of the function

The graph of y = f(x) is the set of all valid pairs (x, f(x)). This makes many function properties visible: intercepts, turning points, asymptotes, intervals of increase or decrease, one-to-one behaviour, and whether a claimed range is plausible.

Graph reading should not be separated from algebra. When an algebraic answer claims that f(x) = 4 has three solutions, the graph should contain three intersections with the horizontal line y = 4.

4. Inverse functions run a relationship backwards

If f maps x to y, then f⁻¹ maps y back to x, provided the mapping is one-to-one on the chosen domain. To find the inverse algebraically:

1. Write y = f(x).
2. Swap x and y.
3. Solve for y.
4. Rename y as f⁻¹(x).
5. State any required domain or range restriction.

For f(x) = 3x − 7, write y = 3x − 7. Swap: x = 3y − 7. Solve: y = (x + 7)/3. Hence f⁻¹(x) = (x + 7)/3.

5. Why x² needs a restricted domain before inversion

The function f(x) = x² on all real x is not one-to-one because f(2) = f(−2) = 4. A backwards machine receiving 4 would not know whether to return 2 or −2. If we restrict the domain to x ≥ 0, the function becomes one-to-one and the inverse is f⁻¹(x) = √x.

An inverse is not created by algebra alone. It is made possible by the structure of the mapping.

6. Inverse graphs reflect in y = x

If (a, b) lies on y = f(x), then (b, a) lies on y = f⁻¹(x). Swapping coordinates is exactly reflection in the line y = x. This gives a geometric test for whether an inverse result makes sense.

7. Composition connects functions in sequence

The composition fg(x) is commonly used to mean f(g(x)) in the school context. The inner function acts first. This order matters.

If f(x) = 2x + 1 and g(x) = x², then

• f(g(x)) = 2x² + 1.
• g(f(x)) = (2x + 1)².

These are not generally equal. Function composition is therefore directional.

8. Worked Example 1 — evaluate a composition

Let f(x) = 3x − 2 and g(x) = 1/(x + 1). Find f(g(2)).

First g(2) = 1/3. Then f(1/3) = 1 − 2 = −1.

The safest approach is to execute the machine chain in order rather than trying to substitute everything mentally at once.

9. Worked Example 2 — find a composite function and its domain

Let f(x) = √x and g(x) = x − 4. Find f(g(x)).

f(g(x)) = √(x − 4). For real values, x − 4 ≥ 0, so the domain is x ≥ 4.

Composition can create a new restriction even when one of the original functions had a much larger domain.

10. Composition and inverse are verification partners

If f and f⁻¹ are genuine inverses on the relevant domains, then

f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.

This provides a clean algebraic check. If the composition does not reduce to x, inspect the inverse derivation and domain assumptions.

11. Graph transformations are coordinate rules

Suppose the original graph is y = f(x). Then common transformations include:

• y = f(x) + a: vertical shift by a.
• y = f(x − a): horizontal shift right by a.
• y = af(x): vertical scale factor |a|, with reflection in the x-axis if a < 0.
• y = f(ax): horizontal scale factor 1/|a|, with reflection in the y-axis if a < 0.
• y = −f(x): reflection in the x-axis.
• y = f(−x): reflection in the y-axis.

The horizontal transformations often feel backwards because the change happens to the input before the function acts.

12. Why f(x − 3) moves the graph right

Suppose the original graph contains the point (2, f(2)). In the transformed graph y = f(x − 3), we need x − 3 = 2 to reproduce the same output, so x = 5. The point therefore moves from x = 2 to x = 5: three units right.

This coordinate reasoning is more reliable than memorising “inside opposite, outside same” without understanding why.

13. Worked Example 3 — transform a quadratic graph

Starting from y = x², describe y = −2(x − 3)² + 5.

• Shift right 3.
• Vertical stretch by factor 2.
• Reflect in the x-axis.
• Shift up 5.

The vertex is therefore (3, 5), and because the coefficient is negative the graph opens downward.

14. Range can often be read after transformation

For y = −2(x − 3)² + 5, the square term is non-negative, so −2(x − 3)² ≤ 0. Hence y ≤ 5. The maximum value is 5.

Graph transformations therefore connect directly to inequalities and range.

15. Worked Example 4 — inverse with a domain restriction

Let f(x) = (x − 1)² + 4 with domain x ≥ 1. Find f⁻¹(x).

Write y = (x − 1)² + 4. Swap x and y:

x = (y − 1)² + 4.

Then (y − 1)² = x − 4. Because the original domain x ≥ 1 corresponds to the positive branch, y − 1 = √(x − 4). Therefore

f⁻¹(x) = 1 + √(x − 4), with domain x ≥ 4.

16. Worked Example 5 — solve using inverse structure

If f(x) = 5x − 8, solve f(x) = 17.

Directly: 5x − 8 = 17, so x = 5. Or use the inverse: f⁻¹(17) = (17 + 8)/5 = 5. The inverse is therefore not merely a new chapter; it packages the act of solving for the input.

17. Common failure patterns

• Confusing f⁻¹(x) with 1/f(x). The inverse function is not the reciprocal.
• Ignoring one-to-one requirements. Algebra may produce two branches when the original domain was not restricted.
• Applying composition in the wrong order. The inner function acts first.
• Forgetting a composite-domain restriction.
• Moving horizontal transformations in the intuitive rather than algebraic direction.
• Changing both x and y coordinates incorrectly during a transformation.
• Using a graph sketch without checking exact intercepts or restrictions.

18. A reliable function routine

1. Identify the function and legal domain.
2. Evaluate inputs carefully, especially expression inputs.
3. For compositions, work from the inside outward.
4. For inverses, check whether the function is one-to-one on the stated domain.
5. Swap variables and solve, preserving the correct branch.
6. Use composition to verify the inverse.
7. For graph transformations, track what changes to the input and what changes to the output.
8. Check range and special points after the transformation.

19. Practice set

1. If f(x) = 4x − 1, find f(3).
2. If f(x) = x² + 2, find f(a − 1).
3. Find the domain of g(x) = 1/(x + 5).
4. Find the inverse of f(x) = 2x + 9.
5. Explain why x² on all real numbers has no inverse function.
6. Given f(x)=x+2 and g(x)=3x, find f(g(x)) and g(f(x)).
7. Find the domain of √(2x − 6).
8. Describe the transformation from y=f(x) to y=f(x−4)+3.
9. Describe y=−f(2x).
10. For f(x)=(x+2)², x≥−2, find f⁻¹(x).
11. Verify by composition that f(x)=3x−5 and f⁻¹(x)=(x+5)/3 are inverses.
12. State the range of y=−3(x−1)²+7.
13. If a point (2,5) lies on y=f(x), where does it move on y=f(x−3)+4?
14. If (a,b) lies on y=f(x), what point lies on y=f⁻¹(x)?
15. Why is the domain of an inverse equal to the range of the original function?
16. If f(x)=1/x and g(x)=x+2, find f(g(x)) and state its domain.
17. For y=f(−x), what happens geometrically?
18. For y=2f(x), what happens to y-coordinates?
19. If f(x)=x² with x≥0, solve f(x)=49 using the inverse.
20. Explain why a composite function can have a smaller domain than either formula appears to suggest at first glance.

Answers

1. 11.
2. (a−1)²+2.
3. x≠−5.
4. f⁻¹(x)=(x−9)/2.
5. Because different inputs x and −x can produce the same output.
6. 3x+2; 3x+6.
7. x≥3.
8. Right 4, up 3.
9. Horizontal scale factor 1/2, then reflection in the x-axis.
10. f⁻¹(x)=√x−2, x≥0.
11. Both compositions simplify to x.
12. y≤7.
13. (5,9).
14. (b,a).
15. Because the inverse receives as inputs exactly the outputs produced by the original.
16. 1/(x+2), x≠−2.
17. Reflection in the y-axis.
18. They are doubled.
19. x=7.
20. Because the output of the inner function must lie inside the domain of the outer function.

20. What mastery looks like

A student has genuine control when function notation feels like input-output logic rather than symbol decoration; when domains and ranges are treated as part of the object; when inverse and composition can be checked rather than guessed; and when graph transformations can be reasoned from coordinates instead of memorised as a disconnected list.

The transfer test is to change representation. Give the student a graph rather than a formula, a restricted quadratic rather than a linear function, or a composition with a hidden domain constraint. If the same input-output reasoning survives, the function system is becoming durable.

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Continue through the BTT Additional Mathematics system

• Additional Mathematics Directory.
• BTT Singapore Mathematics Hub.
• Guide 6: Polynomials, Factor/Remainder Theorem, Partial Fractions and Binomial Expansion.
• Guide 7: Surds, Indices, Inequalities and Exact Algebra.
• Guide 8: Equations, Parameters and Algebraic Proof.