Physics and Mathematics
Composite Function
1. Introduction
In real life, one process often happens after another.
For example:
- First converting temperature from Celsius to Fahrenheit
- Then converting it into Kelvin
In Mathematics, such a situation is handled using a Composite Function.
2. Definition of Composite Function
If two functions f and g are defined such that the range of g is a subset of the domain of f, then the composite function of f and g is denoted by:
[f ∘ g (x) = f(g(x))]
Here:
- g is applied first
- f is applied next
Important Note
Order matters in composite functions.
In general:
[f ∘ g ≠ g ∘ f]
3. Domain of a Composite Function
The domain of [f ∘ g] consists of all x such that:
- [x] belongs to the domain of [g]
- [g(x)] belongs to the domain of [f]
4. Composition of Even and Odd Functions
Important Condition:
The domain of the composite function must be symmetric about 0.
| Outer Function | Inner Function | Composite Function | Reason |
|---|---|---|---|
| Even | Even | Even | [f(g(−x)) = f(g(x))] |
| Even | Odd | Even | [f(−g(x)) = f(g(x))] |
| Odd | Even | Even | [g(−x) = g(x)] |
| Odd | Odd | Odd | [f(−g(x)) = −f(g(x))] |
5. Properties of Composite Functions
| Property | Statement | Mathematical Form |
|---|---|---|
| Non-Commutative | Order of composition matters | [f ∘ g ≠ g ∘ f] |
| Associative | Grouping does not affect result | [f ∘ (g ∘ h) = (f ∘ g) ∘ h] |
| Identity Property | Composition with identity gives same function | [f ∘ I = I ∘ f = f] |
| Domain Restriction | Domain may reduce after composition | Domain of [f ∘ g] ⊆ Domain of g |
| Even-Odd Property | Nature depends on nature of functions | Even ∘ Even = Even |
| Polynomial Closure | Composite of polynomials is polynomial | Polynomial ∘ Polynomial = Polynomial |
| Invertibility | Inverse exists only if both are invertible | [(f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹] |
6. Quick Memory Trick
- Even outside → result always Even
- Odd outside → result depends on inner function
7. Examples with Solutions
Example 1.
If [f(x) = x²] and [g(x) = x + 1], find [f ∘ g (x)].
Solution:
f ∘ g (x) = f(g(x))
= f(x + 1)
= (x + 1)²
Answer:
[f ∘ g (x) = (x + 1)²]
Example 2.
If [f(x) = √x] and [g(x) = x − 2], find [f ∘ g (x)].
Solution:
f ∘ g (x) = √(x − 2)
Domain:
x − 2 ≥ 0 ⇒ x ≥ 2
Example 3.
Find [g ∘ f (x)] for
[f(x) = x²] and [g(x) = x + 1].
Solution:
g ∘ f (x) = g(x²)
= x² + 1
Observation:
[f ∘ g ≠ g ∘ f]
Example 4.
If [f(x) = 2x + 3] and [g(x) = 1 / x], find [f ∘ g (x)].
Solution:
f ∘ g (x) = 2(1 / x) + 3
= (2 + 3x) / x
Domain:
x ≠ 0
Example 5.
If [f(x) = |x|] and [g(x) = x − 4], find [f ∘ g (x)].
Solution:
f ∘ g (x) = |x − 4|
8. Conceptual Questions with Solutions
1. What is meant by a composite function?
A composite function is obtained by applying one function to the result of another function, written as [f(g(x))].
2. Why is the order important in composite functions?
Because changing the order usually gives a different output, hence [f ∘ g ≠ g ∘ f].
3. Can every pair of functions be composed?
No. The range of the inner function must lie within the domain of the outer function.
4. What does f ∘ g mean?
It means first apply g, then apply f to the result.
5. Is composite function always defined?
No. It depends on domain compatibility.
6. Can composite functions be identity functions?
Yes, if [f ∘ g (x) = x], then it becomes an identity function.
7. Is composite of two polynomial functions always a polynomial?
Yes, the composite of two polynomial functions is also a polynomial.
8. What happens to domain after composition?
The domain may get restricted due to the inner function.
9. Is composite of even functions always even?
Yes, if both functions are even, their composite is also even.
10. Can composite functions be inverted?
Yes, if both functions are invertible.
11. Does composite function follow commutative law?
No, composite functions are not commutative.
12. What is the inner function?
The function applied first in composition.
13. What is the outer function?
The function applied after the inner function.
14. Can composite function be constant?
Yes, depending on the functions involved.
15. Why is composite function important?
It helps model multi-step processes in mathematics and real life.
9. FAQs / Common Misconceptions
1. f ∘ g means f × g.
No. It means function composition, not multiplication.
2. g ∘ f is same as f ∘ g.
False. Order matters.
3. Domain remains unchanged after composition.
False. Domain may get restricted.
4. Composite function always exists.
False. Domain-range compatibility is required.
5. Composite of two odd functions is odd.
True, if both are odd.
6. Composite function is always complicated.
Not true. Sometimes it is very simple.
7. Only algebraic functions can be composed.
False. Any functions can be composed.
8. Composite function ignores domain.
Incorrect. Domain is crucial.
9. Composite functions are reversible.
Only if both functions are invertible.
10. Composite functions are not useful.
Wrong. They are extremely important in higher mathematics.
10. Practice Questions with Step-by-Step Solutions
Question 1. If [f(x) = x²] and [g(x) = x − 3], find [f ∘ g (x)].
Solution:
[f ∘ g (x) = f(x − 3)]
[= (x − 3)^2]
Answer:
[f ∘ g (x) = (x − 3)^2]
Question 2. If [f(x) = √x] and [g(x) = x + 4], find [f ∘ g (x)] and its domain.
Solution:
[f ∘ g (x) = √(x + 4)]
[x + 4 ≥ 0] [⇒ x ≥ −4]
Answer:
Domain: [x ≥ −4]
Question 3. Find [g ∘ f (x)] if [f(x) = 2x] and [g(x) = x^2].
Solution:
[g ∘ f (x) ][= (2x)^2 ][= 4x^2]
Question 4. If [f(x) = 1 / x] and [g(x) = x − 1], find [f ∘ g (x)].
Solution:
[f ∘ g (x) = 1 / (x − 1)]
Domain: [x ≠ 1]
Question 5. Find [f ∘ f (x)] if [f(x) = x + 2].
Solution:
[f ∘ f (x) ][= f(x + 2) ][= x + 4]
Question 6. If [f(x) = 2x + 1] and [g(x) = x − 3], find [f ∘ g (x)].
Step-by-Step Solution:
f ∘ g (x) = f(x − 3)
= 2(x − 3) + 1
= 2x − 6 + 1 = 2x − 5
Answer:
[f ∘ g (x) = 2x − 5]
Question 7. If [f(x) = x²] and [g(x) = √x], find [f ∘ g (x)].
Step-by-Step Solution:
f ∘ g (x) = f(√x)
= (√x)² = x
Domain:
x ≥ 0
Conclusion:
Composite function simplifies to x.
Question 8. Find [g ∘ f (x)] if [f(x) = x + 2] and [g(x) = |x|].
Step-by-Step Solution:
g ∘ f (x) = |x + 2|
Answer:
[g ∘ f (x) = |x + 2|]
Question 9. If [f(x) = 1 / x] and [g(x) = x²], find [f ∘ g (x)] and its domain.
Step-by-Step Solution:
f ∘ g (x) = 1 / (x²)
Denominator ≠ 0 ⇒ x ≠ 0
Answer:
[f ∘ g (x) = 1 / x²], domain: [x ≠ 0]
Question 10. Verify whether [f ∘ g = g ∘ f] for [f(x) = x + 1] and [g(x) = x²].
Step-by-Step Solution:
f ∘ g (x) = f(x²) = x² + 1
g ∘ f (x) = g(x + 1) = (x + 1)²
Since [x² + 1 ≠ (x + 1)²]
Conclusion:
[f ∘ g ≠ g ∘ f]
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