Composite & Inverse Functions | DP IB Maths: AA HL Revision Notes 2021

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Composite Functions

What is a composite function?

A composite function is where a function is applied to another function
A composite function can be denoted
- $(f \circ g) (x)$
- $f g (x)$
- $f (g (x))$
The order matters
- $(f \circ g) (x)$ means:
  - First apply g to x to get $g (x)$
  - Then apply f to the previous output to get $f (g (x))$
  - Always start with the function closest to the variable
- $(f \circ g) (x)$ is not usually equal to $(g \circ f) (x)$

How do I find the domain and range of a composite function?

The domain of $f \circ g$ is the set of values of $x$ ...
- which are a subset of the domain of g
- which maps g to a value that is in the domain of f
The range of $f \circ g$ is the set of values of $x$ ...
- which are a subset of the range of f
- found by applying f to the range of g
To find the domain and range of $f \circ g$
- First find the range of g
- Restrict these values to the values that are within the domain of f
  - The domain is the set of values that produce the restricted range of g
  - The range is the set of values that are produced using the restricted range of g as the domain for f
For example: let $f (x) = 2 x + 1, - 5 \leq x \leq 5$ and $g (x) = \sqrt{x}, 1 \leq x \leq 49$
- The range of g is $1 \leq g (x) \leq 7$
  - Restricting this to fit the domain of f results in $1 \leq g (x) \leq 5$
- The domain of $f \circ g$ is therefore $1 \leq x \leq 25$
  - These are the values of x which map to $1 \leq g (x) \leq 5$
- The range of $f \circ g$ is therefore $3 \leq (f \circ g) (x) \leq 11$
  - These are the values which f maps $1 \leq g (x) \leq 5$ to

Examiner Tip

Make sure you know what your GDC is capable of with regard to functions
- You may be able to store individual functions and find composite functions and their values for particular inputs
- You may be able to graph composite functions directly and so deduce their domain and range from the graph
The link between the domains and ranges of a function and its inverse can act as a check for your solution
$f f (x)$ is not the same as ${[f (x)]}^{2}$

Worked example

Given $f (x) = \sqrt{x + 4}$ and $g (x) = 3 + 2 x$ :

Write down the value of

(g \circ f) (12)

2-3-2-ib-aa-sl-composite-functions-a-we-solution

Write down an expression for

(f \circ g) (x)

2-3-2-ib-aa-sl-composite-functions-b-we-solution

Write down an expression for

(g \circ g) (x)

2-3-2-ib-aa-sl-composite-functions-c-we-solution

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Inverse Functions

What is an inverse function?

Only one-to-one functions have inverses
A function has an inverse if its graph passes the horizontal line test
- Any horizontal line will intersect with the graph at most once
The identity function $id$ maps each value to itself
- $id (x) = x$
If $f \circ g$ and $g \circ f$ have the same effect as the identity function then $f$ and $g$ are inverses
Given a function $f (x)$ we denote the inverse function as $f^{- 1} (x)$
An inverse function reverses the effect of a function
- $f (2) = 5$ means $f^{- 1} (5) = 2$
Inverse functions are used to solve equations
- The solution of $f (x) = 5$ is $x = f^{- 1} (5)$
A composite function made of $f$ and $f^{- 1}$ has the same effect as the identity function
- $(f \circ f^{- 1}) (x) = (f^{- 1} \circ f) (x) = x$

What are the connections between a function and its inverse function?

The domain of a function becomes the range of its inverse
The range of a function becomes the domain of its inverse
The graph of $y = f^{- 1} (x)$ is a reflection of the graph $y = f (x)$ in the line $y = x$
- Therefore solutions to $f (x) = x$ or $f^{- 1} (x) = x$ will also be solutions to $f (x) = f^{- 1} (x)$
  - There could be other solutions to $f (x) = f^{- 1} (x)$ that don't lie on the line $y = x$

Inverse Functions Notes Diagram 2

How do I find the inverse of a function?

STEP 1: Swap the x and y in $y = f (x)$
- If $y = f^{- 1} (x)$ then $x = f (y)$
STEP 2: Rearrange $x = f (y)$ to make $y$ the subject
Note this can be done in any order
- Rearrange $y = f (x)$ to make $x$ the subject
- Swap $x$ and $y$

Can many-to-one functions ever have inverses?

You can restrict the domain of a many-to-one function so that it has an inverse
Choose a subset of the domain where the function is one-to-one

The inverse will be determined by the restricted domain
Note that a many-to-one function can only have an inverse if its domain is restricted first

For quadratics – use the vertex as the upper or lower bound for the restricted domain

For $f (x) = x^{2}$ restrict the domain so 0 is either the maximum or minimum value

For example: $x \geq 0$ or $x \leq 0$

For $f (x) = a {(x - h)}^{2} + k$ restrict the domain so h is either the maximum or minimum value

For example: $x \geq h$ or $x \leq h$

For trigonometric functions – use part of a cycle as the restricted domain

For $f (x) = \sin x$ restrict the domain to half a cycle between a maximum and a minimum

For example: $- \frac{π}{2} \leq x \leq \frac{π}{2}$

For $f (x) = \cos x$ restrict the domain to half a cycle between maximum and a minimum

For example: $0 \leq x \leq π$

For $f (x) = \tan x$ restrict the domain to one cycle between two asymptotes

For example: $- \frac{π}{2} < x < \frac{π}{2}$

How do I find the inverse function after restricting the domain?

The range of the inverse is the same as the restricted domain of the original function
The inverse function is determined by the restricted domain

Restricting the domain differently will change the inverse function

Use the range of the inverse to help find the inverse function

Restricting the domain of $f (x) = x^{2}$ to $x \geq 0$ means the range of the inverse is $f^{- 1} (x) \geq 0$

Therefore $f^{- 1} (x) = \sqrt{x}$

Restricting the domain of $f (x) = x^{2}$ to $x \leq 0$ means the range of the inverse is $f^{- 1} (x) \leq 0$

Therefore $f^{- 1} (x) = - \sqrt{x}$

Examiner Tip

Remember that an inverse function is a reflection of the original function in the line $y = x$
- Use your GDC to plot the function and its inverse on the same graph to visually check this
$f^{- 1} (x)$ is not the same as $\frac{1}{f (x)}$

Worked example

The function $f (x) = {(x - 2)}^{2} + 5, x \leq m$ has an inverse.

Write down the largest possible value of

m

d7t4IIb~_2-3-2-ib-aa-hl-inverse-functions-a-we-solution

Find the inverse of

f (x)

2-3-2-ib-aa-hl-inverse-functions-b-we-solution

Find the domain of

f^{- 1} (x)

2-3-2-ib-aa-hl-inverse-functions-c-we-solution

Find the value of

k

such that

f (k) = 9

2-3-2-ib-aa-hl-inverse-functions-d-we-solution

Composite & Inverse Functions (DP IB Maths: AA HL)

Revision Note

Composite Functions

What is a composite function?

How do I find the domain and range of a composite function?

Examiner Tip

Worked example

Inverse Functions

What is an inverse function?

What are the connections between a function and its inverse function?

How do I find the inverse of a function?

Can many-to-one functions ever have inverses?

How do I find the inverse function after restricting the domain?

Examiner Tip

Worked example

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Author: Dan