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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

◆ 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 $

Note:

\( fog \neq gof \).

 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

Consider the functions $f(x) = 4x – 6$ and $g(x) = 2x – 1$.

a. The function $r(x)$ is defined as $r(x) = f(x^2)$. Define $r(x)$.

b. Using the results of the previous part, define $q(x)$, which is $g(f(x^2))$.

▶️Answer/Explanation

Solution:

a.

Replace each occurrence of $x$ in $f(x) = 4x – 6$ with $x^2$.

$
r(x) = 4x^2 – 6
$

b.

In the previous part, $f(x^2) = 4x^2 – 6$.

This now needs to be substituted into $g(x)$ wherever we see $x$ occur.

Replace each occurrence of $x$ in $g(x) = 2x – 1$ with $4x^2 – 6$:

$
q(x) = 2(4x^2 – 6) – 1
$

$
q(x) = 8x^2 – 12 – 1
$

$
q(x) = 8x^2 – 13
$

◆ 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

Do the following conversion

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 \( f(x) = 1 – 2x \) 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., \( f(x) = \frac{1}{x} \)).

Domain Restriction in Inverse Functions

A function must be one-to-one to have an inverse that is also a function.
If a function is many-to-one, its inverse is not valid as a function (just a relation).
However, we can restrict the domain of the original function to make it one-to-one, thereby allowing the inverse to be a valid function.
To make an inverse valid, restrict the domain of the original function so that it passes the horizontal line test.

 INVERSE FUNCTION (with domain restriction)

Example:

Find the inverse of \( f(x) = x^2 + 2x – 3 \), by restricting the domain.

▶️Answer/Explanation

Solution:

Completed square form:

\( f(x) = x^2 + 2x – 3 = (x + 1)^2 – 4 \)

Let \( y = (x + 1)^2 – 4 \), then switch \( x \) and \( y \):

\( x = (y + 1)^2 – 4 \)

Solving for \( y \):

\( x + 4 = (y + 1)^2 \)
\( \sqrt{x + 4} = y + 1 \)
\( y = -1 \pm \sqrt{x + 4} \)

This is not a function due to the “±”.

To make it a function, restrict the domain of the original function to \( x \ge -1 \). Then take the positive square root only:

\( f^{-1}(x) = -1 + \sqrt{x + 4} \),   for \( x \ge -4 \)

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