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IB Mathematics AI AHL Composite functions f∘g MAI Study Notes - New Syllabus

IB Mathematics AI AHL Composite functions f∘g MAI Study Notes

LEARNING OBJECTIVE

  • Composite functions in context.

Key Concepts: 

  • Composite & Inverse Functions

MAI HL and SL Notes – All topics

COMPOSITION OF FUNCTIONS: fog

◆ Consider the function \( f(x) = x^2 \).

Notice that:
\( f(5) = 5^2 \)
\( f(a) = a^2 \)
\( f(3a + 5) = (3a + 5)^2 \)
\( f(3x + 5) = (3x + 5)^2 \)

In the last case, the input value for \( f \) is another function of \( x \). This combines two functions:
 \( f(x) = x^2 \)
\( g(x) = 3x + 5 \)

to create a new function \( y = (3x + 5)^2 \). This new function is denoted by \( fog \).

◆ DEFINITION
For two functions \( f \) and \( g \), the composite function \( fog \) is defined by:
$ (fog)(x) = f(g(x))$

The operation is called composition. We say \( fog \) is the composite function of \( f \) and \( g \). 

Example:
For \( f(x) = x^2 \) and \( g(x) = 3x + 5 \):
$ (fog)(x) = f(g(x)) = f(3x + 5) = (3x + 5)^2 $

Similarly, the composite function \( gof \) is:
$ (gof)(x) = g(f(x)) = g(x^2) = 3x^2 + 5 $

Notice:
\( fog \neq gof \).
 No need to expand; just plug \( g \) into \( f \) for \( fog \), and vice versa for \( gof \).

For three functions \( f(x) = x^2 \), \( g(x) = 3x + 5 \), \( h(x) = \sqrt{x} \):
$ (fogoh)(x) = f(g(h(x))) = f(3\sqrt{x} + 5) = (3\sqrt{x} + 5)^2 $
This shows \( (fogoh) = (fog)oh \). 

Example

Let \( f(x) = 2x^2 – 1 \) and \( g(x) = x + 1 \). Find:
(a) \( (fog)(x) \)
(b) \( (gof)(x) \)
(c) \( (fog)(1) \)
(d) \( (gof)(1) \)

▶️Answer/Explanation

Solution:

(a) \( (fog)(x) = 2(x + 1)^2 – 1 \)
(b) \( (gof)(x) = (2x^2 – 1) + 1 = 2x^2 \)
(c) \( (fog)(1) = 7 \)
(d) \( (gof)(1) = 2 \)

Alternatively:
(c) \( (fog)(1) = f(g(1)) = f(2) = 7 \)
(d) \( (gof)(1) = g(f(1)) = g(1) = 2 \) 

Example

Let \( f(x) = \frac{x + 1}{2} \) and \( g(x) = \sqrt{x} \). Find:
(a) \( (fog)(x) \)
(b) \( (gof)(x) \)
(c) \( (fof)(x) \)
(d) \( (gog)(x) \)
(e) \( (fofof)(x) \) in two ways: as \( fo(fof) \) and as \( (fof)of \).

▶️Answer/Explanation

Solution:

(a) \( (fog)(x) = \frac{\sqrt{x} + 1}{2} \)
(b) \( (gof)(x) = \sqrt{\frac{x + 1}{2}} \)
(c) \( (fof)(x) = \frac{\frac{x + 1}{2} + 1}{2} = \frac{x + 3}{4} \)
(d) \( (gog)(x) = \sqrt{\sqrt{x}} = \sqrt[4]{x} \)
(e) \( (fofof)(x) = \frac{\frac{x + 3}{4} + 1}{2} = \frac{x + 7}{8} \) (same for both methods).

◆ THE IDENTITY FUNCTION \( i(x) \)

It maps \( x \) to itself:
$ i(x) = x \quad \text{or} \quad i: x \mapsto x $

Properties:
\( (foi)(x) = f(x) \)
 \( (iof)(x) = f(x) \)

◆ COMPOSITION IN REAL LIFE PROBLEMS  

Example:
The area of a square of side \( a \) is \( A = a^2 \).
For a square of side \( 2b + 3 \), the area is \( (2b + 3)^2 \).

Here:
\( f(x) = x^2 \)
\( g(x) = 2x + 3 \)
\( (fog)(x) = (2x + 3)^2 \) 

Example

Temperature conversions:
 \( F = 1.8C + 32 \) (Celsius to Fahrenheit)
 \( C = K – 273.15 \) (Kelvin to Celsius)

▶️Answer/Explanation

Solution:

Combining:
$ F = 1.8(K – 273.15) + 32 = 1.8K – 459.67 $

This is analogous to:
 \( f(x) = 1.8x + 32 \)
 \( g(x) = x – 273.15 \)
 \( (fog)(x) = 1.8(x – 273.15) + 32 \).

THE INVERSE FUNCTION: \( f^{-1} \)

◆ DEFINITION
For \( f: \mathbb{R} \rightarrow \mathbb{R} \), the inverse function \( f^{-1} \) satisfies:
$ f(x) = y \iff f^{-1}(y) = x $

Steps to find \( f^{-1} \):
1. Set \( f(x) = y \).
2. Solve for \( x \).
3. Replace \( y \) with \( x \). 

Example:
For \( f(x) = x + 10 \):
1. \( x + 10 = y \)
2. \( x = y – 10 \)
3. \( f^{-1}(x) = x – 10 \)

Properties:
\( (f^{-1})^{-1} = f \)
 \( D_{f^{-1}} = R_f \) and \( R_{f^{-1}} = D_f \)
 \( (fof^{-1})(x) = x \) and \( (f^{-1}of)(x) = x \) 

Example

Let \( f(x) = 2x^2 – 1 \) (\( x \geq 0 \)). Find:
(a) \( f^{-1}(x) \)
(b) \( f^{-1}(49) \)

▶️Answer/Explanation

Solution:

(a)
1. \( 2x^2 – 1 = y \)
2. \( x = \sqrt{\frac{y + 1}{2}} \)
3. \( f^{-1}(x) = \sqrt{\frac{x + 1}{2}} \)

(b) \( f^{-1}(49) = 5 \). 

Example

Let \( f(x) = \frac{x + 1}{x + 2} \).
(a) Show \( f^{-1}(x) = \frac{2x – 1}{1 – x} \).
(b) Verify \( (fof^{-1})(x) = x \).

▶️Answer/Explanation

Solution:

(a) Solve \( \frac{x + 1}{x + 2} = y \) to get \( x = \frac{2y – 1}{1 – y} \).
(b) Substitute and simplify to confirm. 

Example

Let \( f(x) = 1 – 2x \) and \( g(x) = \frac{1}{x} \). Find:
(a) \( (fog)(x) \)
(b) \( (gof)(x) \)
(c) \( (gof^{-1})(x) \)
(d) \( (fog^{-1})(x) \)
(e) \( (fog)^{-1}(x) \)
(f) \( (f^{-1}og^{-1})(x) \)

▶️Answer/Explanation

Solution:

(a) \( (fog)(x) = 1 – \frac{2}{x} \)
(b) \( (gof)(x) = \frac{1}{1 – 2x} \)
(c) \( f^{-1}(x) = \frac{1 – x}{2} \), so \( (gof^{-1})(x) = \frac{2}{1 – x} \)
(d) \( g^{-1}(x) = \frac{1}{x} \), so \( (fog^{-1})(x) = 1 – \frac{2}{x} \)
(e) \( (fog)^{-1}(x) = \frac{2}{1 – x} \)
(f) \( (f^{-1}og^{-1})(x) = \frac{1 – \frac{1}{x}}{2} = \frac{x – 1}{2x} \)

Notice:
 \( (fog)^{-1} = g^{-1}of^{-1} \).
A function is self-inverse if \( f^{-1} = f \) (e.g., \( f(x) = \frac{1}{x} \)).

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