Functional equations path · E01

Before this lesson: Mathematical language and sets, Equations and quadratic tools

Your goal: Use function properties precisely without assuming a formula.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Names you may know: injective and surjective functions; inverse functions.

Functions: Inverse, Composition and One-to-One: the key idea

A function f:A→B assigns exactly one output in B to each input in A. Its range may be smaller than its codomain B. It is injective if equal outputs force equal inputs, and surjective if every element of B occurs as an output. A two-sided inverse exists exactly for a bijection. Composition f∘g requires outputs of g to lie in the domain of f. A formula alone does not settle these properties: x² is bijective on [0,∞)→[0,∞), but not injective on ℝ→ℝ. Continuity is an additional hypothesis, not a property of every function.

A worked example

Is f:ℝ→ℝ, f(x)=2x+3 bijective?

If f(a)=f(b), then 2a+3=2b+3, so a=b: injective. For any real y, x=(y−3)/2 is real and f(x)=y: surjective. The inverse is f⁻¹(y)=(y−3)/2.

Your turn: change one thing

Is g:ℤ→ℤ, g(n)=2n surjective?

Try this on paper before opening the explanation.

Compare your reasoning

No: no integer input maps to an odd integer such as 1. It is injective, because 2a=2b implies a=b.

Pause and check

A trap to avoid: Confusing codomain with range or assuming every function is continuous.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove that the composition of two injective functions is injective whenever the composition is defined.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Let f(g(a))=f(g(b)). Injectivity of f gives g(a)=g(b); injectivity of g then gives a=b. This is precisely injectivity of f∘g.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. The range of a function consists of what?

Foundation

  1. Outputs actually attained
  2. All real numbers always
  3. Only inputs
  4. Every element of its domain
Hint

Distinguish possible target values from achieved values.

Answer and reasoning

Outputs actually attained. The range is {f(x):x belongs to the domain}.

2. Injectivity means what?

Foundation

  1. Every codomain value is attained
  2. All inputs have equal outputs
  3. The function is continuous
  4. Equal outputs imply equal inputs
Hint

Use the one-to-one definition.

Answer and reasoning

Equal outputs imply equal inputs. f(a)=f(b) must force a=b.

3. The inverse of f(x)=2x+3 on ℝ is what?

Core

  1. 1/(2x+3)
  2. (x−3)/2
  3. 2x−3
  4. (x+3)/2
Hint

Solve y=2x+3 for x.

Answer and reasoning

(x−3)/2. x=(y−3)/2; an inverse function is not a reciprocal.

4. Is n↦2n from ℤ to ℤ surjective?

Core

  1. No, zero is missed
  2. Only for positive n
  3. No, odd integers are missed
  4. Yes
Hint

Test the target 1.

Answer and reasoning

No, odd integers are missed. 2n=1 has no integer solution.

5. Why is x↦x² not injective on ℝ?

Stretch

  1. It is not continuous
  2. 1 and −1 have the same output
  3. It has no outputs
  4. It is negative
Hint

Find two different inputs.

Answer and reasoning

1 and −1 have the same output. Both 1² and (−1)² equal 1.

6. A two-sided inverse exists exactly when a function is what?

Stretch

  1. Constant
  2. Bijective
  3. Only injective
  4. Only continuous
Hint

Both one-to-one and onto are needed.

Answer and reasoning

Bijective. An inverse must assign each codomain output its unique original input.

7. Surjective means what?

Foundation

  1. The graph is continuous
  2. Every codomain value is attained
  3. Every input is positive
  4. All outputs are equal
Hint

Surjective is also called onto.

Answer and reasoning

Every codomain value is attained. For each target y there must be an allowed x with f(x)=y.

8. For f(x)=x+2 and g(x)=3x on ℝ, what is f(g(4))?

Core

  1. 6
  2. 14
  3. 18
  4. 12
Hint

Apply g first.

Answer and reasoning

14. g(4)=12, then f(12)=14.

9. Can a function’s bijectivity change when its domain changes?

Stretch

  1. Only for constants
  2. Yes
  3. No
  4. Only if its formula also changes
Hint

Compare x² on ℝ and on [0,∞).

Answer and reasoning

Yes. Restricting the domain can remove collisions or change which outputs occur.

Choose your next step

Continue to Solving functional equations. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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