Patrick's copy of Inverse and composite functions

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Patrick Joyce

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Patrick Joyce 7 years, 8 months ago

Created this as a copy of Inverse and composite functions.

Name	Status	Author	Last Modified
Inverse and composite functions	Ready to use	Bradley Bush	20/11/2019 14:40
Patrick's copy of Inverse and composite functions	draft	Patrick Joyce	27/07/2017 15:58
Inverse and composite functions [L3]	draft	Abbi Mullins	24/07/2018 14:02
Maria's copy of Inverse and composite functions	Needs to be tested	Maria Aneiros	05/02/2020 06:59
Inverse and composite functions	draft	Xiaodan Leng	11/07/2019 01:46

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Question statement

Given a function $f(x)$ , the inverse function $f^{-1}(x)$ reverses whatever $f$ does.

Function composition is applying one function to the results of another.

The following questions will ask you to find the inverses and compositions of some functions.

Give all of your answers in terms of $x$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

			a	list	[ 2, 4, 2, 3, 2 ]
			b	list	[ 9, 7, 5, 7, 6 ]

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
repeat(random(2..5),5)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `list`

[ 2, 4, 2, 3, 2 ]

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Parts

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Part f) Choose one from a list

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Ask the student a question, and give any hints about how they should answer this part.

Find $f^{-1}(x)$ when $\simplify{f(x)={a[0]}x+{a[1]} }$ .

$\displaystyle f^{-1}(x)=$ Unnamed gap

Advice

a)

$f^{-1}(x)$ is the function with the property that $f^{-1}(f(x)) = x$ .

To find this, we first set $x=f(y)$ and rearrange to find $y$ in terms of $x$ , i.e. $y = f^{-1}(x)$ .

$\begin{align} f(y)=\simplify{{a[0]}y+{a[1]}}&=x\\ \simplify{{a[0]}y}&=x-\var{a[1]}\\[1em] y&=\simplify[]{(x-{a[1]})/{a[0]}}\\[1em] f^{-1}(x)&=\simplify{(x-{a[1]})/{a[0]}}\text{.}\\ \end{align}$

b)

We use the same method as part a) to find $g^{-1}(x)$ :

$\begin{align} g(y)=\simplify{{a[3]}y-{a[2]}}&=x\\ \simplify{{a[3]}y}&=x+\var{a[2]}\\[1em] y&=\simplify[]{(x+{a[2]})/{a[3]}}\\[1em] g^{-1}(x)&=\simplify{(x+{a[2]})/{a[3]}}\text{.}\\ \end{align}$

c)

$(g^{-1} \circ f^{-1})(x)$ is the function which first applies $f^{-1}(x)$ and then applies $g^{-1}$ to the result of that.

We use the previous answers: $f^{-1}(x)=\simplify{(x-{a[1]})/{a[0]}}$ and $g^{-1}(x)=\simplify{(x+{a[2]})/{a[3]}}$ to find the definition of $(g^{-1} \circ f^{-1})(x)$ by substituting $f^{-1}(x)$ everywhere $x$ occurs in the definition of $g^{-1}(x)$ .

$\begin{align} (g^{-1}\circ f^{-1}) (x)&=g^{-1}(f^{-1}(x))\\[1em] &=g^{-1} \left( \simplify[]{(x-{a[1]})/{a[0]}} \right) \\[1em] &=\frac{\left(\simplify[]{(x-{a[1]})/{a[0]}}\right)+\simplify[]{{a[2]}}}{\var{a[3]}\text{.}} \end{align}$

d)

$(f \circ g)(x)$ is the function which first applies $g(x)$ and then applies $f$ to the result of that.

We find the definition of $(f \circ g)(x)$ by substituting $g(x)$ everywhere that $x$ occurs in the definition of $f(x)$ .

$\begin{align} (f\circ g)(x)&=f(g(x))\\ &=f(\simplify{{a[3]}x-{a[2]}})\\ &=\simplify{{a[0]}({a[3]}x-{a[2]})+{a[1]}} \end{align}$

e)

Now that we have the definition of $(f \circ g)(x)$ , we can find its inverse by using the same method as in parts a) and b).

$\begin{align} (f \circ g)(y) &= x \\ \simplify{{a[0]}({a[3]}y-{a[2]})+{a[1]}}&=x\\ \simplify{{a[0]}({a[3]}y-{a[2]})}&=x-\var{a[1]}\\[1em] \simplify{{a[3]}y-{a[2]}}&=\frac{(x-\var{a[1]})}{\var{a[0]}}\\[1em] \simplify{{a[3]}y}&=\left( \frac{(x-\var{a[1]})}{\var{a[0]}} \right) +\var{a[2]}\\[1em] y&=\frac{\left(\frac{(x-\var{a[1]})}{\var{a[0]}}\right)+\var{a[2]})}{\var{a[3]}}\\[1em] (f\circ g)^{-1}(x)&=\frac{\left(\frac{(x-\var{a[1]})}{\var{a[0]}}\right)+\var{a[2]}}{\var{a[3]}}\\[1em] \end{align}$

We can see that in this case $(f\circ g)^{-1}(x) = (g^{-1}\circ f^{-1}) (x)$ .

f)

So long as the inverses of $f$ and $g$ exist and they can be composed, it is always true that

$(f \circ g)^{-1}(x) \equiv (g^{-1} \circ f^{-1}) (x)\text{.}$

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Patrick's copy of Inverse and composite functions

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Patrick Joyce 7 years, 8 months ago

Error in variable testing condition

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Generated value: `list`

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

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Error

Patrick's copy of Inverse and composite functions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Patrick Joyce 7 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

Generated value: `list`