Lois's copy of Inverse and composite functions

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Lois Rollings

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Lois Rollings 7 years, 5 months ago

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Question statement

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ with $\simplify{f(x)={a[0]}x+{a[1]} }$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ with $\simplify{g(x)={a[3]}x-{a[2]} }$ .

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

Give any introductory information the student needs.

			Name	Type	Generated Value

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

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`

[ 4, 2, 3, 2, 2 ]

→ Used by:

Statement
Advice
"Part a)" → "Unnamed gap" - Correct answer
"Part b)" → "Unnamed gap" - Correct answer
"Part c)" → "Unnamed gap" - Correct answer
"Part d)" → "Unnamed gap" - Correct answer
"Part e)" → "Unnamed gap" - Correct answer

Parts

Part a) Gap-fill
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Part b) Gap-fill
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Part c) Gap-fill
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Part d) Gap-fill
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Part e) Gap-fill
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Part f) Choose one from a list

Gap-fill

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

Find $f^{-1}: \mathbb{R} \rightarrow \mathbb{R}$ .

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

Advice

a)

$f^{-1} : \mathbb{R} \rightarrow \mathbb{R}$ .

To find $f^{-1}(x)$ , we can 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)

$g^{-1} : \mathbb{R} \rightarrow \mathbb{R}$ .

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)

$g^{-1} \circ f^{-1} : \mathbb{R} \rightarrow \mathbb{R}$ .

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

$(f \circ g) : \mathbb{R} \rightarrow \mathbb{R}$ .

$(f \circ g)(x)$ 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)

$(f \circ g)^{-1} : \mathbb{R} \rightarrow \mathbb{R}$ .

We can use the same method as above to find $(f \circ g)^{-1}(x)$ .

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

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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This question is used in the following exams:

Mappings Quiz by Lois Rollings in Algebra 1 - Possibilities.
Mappings Quiz by Lois Rollings in Algebra 1 - 2020.

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Lois's copy of Inverse and composite functions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Lois Rollings 7 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: `list`

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

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Error

Lois's copy of Inverse and composite functions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Lois Rollings 7 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

Generated value: `list`