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

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17 July 2023

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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 $(g \circ f) (x)$

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

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
$f f (x)$ is not the same as ${[f (x)]}^{2}$

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

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

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 $y$

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 $\frac{1}{f (x)}$