Introduction to Functions

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Simon Thomas

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Simon Thomas 5 years, 2 months ago

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Simon Thomas 5 years, 2 months ago

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Name	Status	Author	Last Modified
Introduction to Functions	draft	Simon Thomas	24/02/2020 13:55
Composite Functions	draft	Simon Thomas	24/02/2020 13:55

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Question statement

You are given the following functions:

$f(x) = \var{a}x+\var{b}$

$g(x) = \var{c}x-\var{d}$

$h(x) = \var{d}x^2-\var{b}x+\var{c-1}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	6
b	integer	4
c	integer	4
d	integer	3
x	integer	3

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

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

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

Statement
Advice
"Unnamed part" → "gap 1" - Minimum accepted value
"Unnamed part" → "gap 1" - Maximum accepted value
"Unnamed part" → "gap 4" - Minimum accepted value
"Unnamed part" → "gap 4" - Maximum accepted value

Parts

Gap-fill
1. Gap gap 1 Number entry
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2. Gap gap 2 Number entry
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3. Gap gap 3 Number entry
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4. Gap gap 4 Number entry
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5. Gap gap 5 Number entry
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6. Gap gap 6 Number entry
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7. Gap gap 7 Number entry
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Gap-fill

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

Calculate:

(i) $f(\var{x})=$ gap 1

(ii) $h(\var{x+2})=$ gap 2

(iii) $g(x^2)=$ gap 3 given that $x=\var{x}$

(iv) $ff(\var{x})=$ gap 4

(v) $gh(\var{x+2})=$ gap 5

(vi) $hg(\var{x+2})=$ gap 6

(vii) $h(\var{x+2})g(\var{x+2})=$ gap 7

Advice

(i) To find $f(\var{x})$ we simply substitute $x=\var{x}$ into our function $f(x) = \var{a}x+\var{b}$

Hence $f(\var{x})=\var{a}(\var{x})+\var{b}=\var{f(x)}$

(ii) To find $h(\var{x+2})$ we simply substitute $x=\var{x+2}$ into our function $h(x) = \var{d}x^2-\var{b}x+\var{c-1}$

Hence $h(\var{x+2})=\var{d}(\var{x+2})^2-\var{b}(\var{x+2})+\var{c-1}=\var{h(x+2)}$

(iii) To find $g(x^2)$ given that $x=\var{x}$ we note that $x^2=\var{x}^2=\var{x^2}$

Hence we just need to calculate $g(\var{x^2})= \var{c}(\var{x^2})-\var{d}=\var{g(x^2)}$

(iv) From (i) we note that $f(\var{x})=\var{f(x)}$ and hence $ff(\var{x})=f(f(\var{x}))=f(\var{f(x)})=\var{a}(\var{f(x)})+\var{b}=\var{f(f(x))}$

(v) We note that $gh(\var{x+2})=g(h(\var{x+2}))=g(\var{h(x+2)})$ using our answer from (ii)

But then $g(\var{h(x+2)}) = \var{c}(\var{h(x+2)})-\var{d}=\var{g(h(x+2))}$

(vi) Note that $hg(\var{x+2})=h(g(\var{x+2}))$

So first we must calculate $g(\var{x+2})=\var{c}(\var{(x+2)})-\var{d}=\var{g(x+2)}$

Then $hg(\var{x+2}) = h(\var{g(x+2)}) = \var{d}(\var{g(x+2)})^2-\var{b}(\var{g(x+2)})+\var{c-1} =\var{h(g(x+2))}$

Note the difference between (v) and (vi):

To find $hg(\var{x+2})$ we first find $g(\var{x+2})$ , then apply $h$ to this answer

To find $gh(\var{x+2})$ we first find $h(\var{x+2})$ , then apply $g$ to this answer

(vii) $h(\var{x+2})g(\var{x+2})$ simply means $h(\var{x+2}) \times g(\var{x+2})=\var{h(x+2)} \times \var{g(x+2)}=\var{h(x+2)*g(x+2)}$

Give a worked solution to the whole question.

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This question is used in the following exam:

Function notation by Simon Thomas in Maths support.

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Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Introduction to Functions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`