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GCSE/IGCSE-FM Functions

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1 GCSE/IGCSE-FM Functions
Dr J Frost Last modified: 18th June 2017

2 OVERVIEW #1: Understanding of functions #2: Inverse Functions GCSE
IGCSEFM GCSE #3: Composite Functions GCSE

3 OVERVIEW #4: Piecewise functions
#5: Domain/Range of common functions (particularly quadratic and trigonometric) IGCSEFM IGCSEFM #6: Domain/Range of other functions #7: Constructing a function based on a given domain/range. IGCSEFM IGCSEFM

4 Name of the function (usually 𝑓 or 𝑔)
What are Functions? A function is something which provides a rule on how to map inputs to outputs. From primary school you might have seen this as a β€˜number machine’. Input Output f π‘₯ 2π‘₯ ? Input Output Name of the function (usually 𝑓 or 𝑔) 𝑓(π‘₯)=2π‘₯

5 𝑓(π‘₯)=π‘₯2+2 Check Your Understanding ? ? ? ? What does this function do?
It squares the input then adds 2 to it. Q1 ? What is 𝑓(3)? 𝒇 πŸ‘ = πŸ‘ 𝟐 +𝟐=𝟏𝟏 What is 𝑓(βˆ’5)? 𝒇 βˆ’πŸ“ = βˆ’πŸ“ 𝟐 +𝟐=πŸπŸ• If 𝑓 π‘Ž =38, what is π‘Ž? 𝒂 𝟐 +𝟐=πŸ‘πŸ– So 𝒂=Β±πŸ” Q2 ? Q3 ? This question is asking the opposite, i.e. β€œwhat input π‘Ž would give an output of 38?” Q4 ?

6 Algebraic Inputs If 𝑓 π‘₯ =π‘₯+1 what is: If 𝑓 π‘₯ = π‘₯ 2 βˆ’1 what is: ? ? ? ?
If you change the input of the function (π‘₯), just replace each occurrence of π‘₯ in the output. If 𝑓 π‘₯ =π‘₯+1 what is: If 𝑓 π‘₯ = π‘₯ 2 βˆ’1 what is: 𝑓 π‘₯βˆ’1 = π’™βˆ’πŸ +𝟏=𝒙 𝑓 π‘₯ 2 = 𝒙 𝟐 +𝟏 𝑓 π‘₯ 2 = 𝒙+𝟏 𝟐 𝑓 2π‘₯ =πŸπ’™+𝟏 ? 𝑓 π‘₯βˆ’1 = π’™βˆ’πŸ 𝟐 βˆ’πŸ = 𝒙 𝟐 βˆ’πŸπ’™ 𝑓 2π‘₯ = πŸπ’™ 𝟐 βˆ’πŸ =πŸ’ 𝒙 𝟐 βˆ’πŸ 𝑓 π‘₯ 2 +1 = 𝒙 𝟐 +𝟏 𝟐 βˆ’πŸ = 𝒙 πŸ’ +𝟐 𝒙 𝟐 ? ? ? ? ? ? If 𝑓 π‘₯ =2π‘₯ what is: 𝑓 π‘₯βˆ’1 =𝟐 π’™βˆ’πŸ =πŸπ’™βˆ’πŸ 𝑓 π‘₯ 2 =𝟐 𝒙 𝟐 𝑓 π‘₯ 2 = πŸπ’™ 𝟐 =πŸ’ 𝒙 𝟐 ? ? ?

7 Test Your Understanding
If 𝑔 π‘₯ =3π‘₯βˆ’1, determine: 𝑔 π‘₯βˆ’1 =πŸ‘ π’™βˆ’πŸ βˆ’πŸ=πŸ‘π’™βˆ’πŸ’ 𝑔 2π‘₯ =πŸ‘ πŸπ’™ βˆ’πŸ=πŸ”π’™βˆ’πŸ 𝑔 π‘₯ 3 =πŸ‘ 𝒙 πŸ‘ βˆ’πŸ ? ? ? B If 𝑓 π‘₯ =2π‘₯+1, solve 𝑓 π‘₯ 2 =51 2 π‘₯ 2 +1=51 π‘₯=Β±5 ?

8 Exercise 1 ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet)
If 𝑓 π‘₯ =2π‘₯+5, find: 𝑓 3 =𝟏𝟏 𝑓 βˆ’1 =πŸ‘ 𝑓 =πŸ” If 𝑓 π‘₯ = π‘₯ 2 +5, find 𝑓 βˆ’1 =πŸ” the possible values of π‘Ž such that 𝑓 π‘Ž = 𝒂=Β±πŸ” The possible values of π‘˜ such that 𝑓 π‘˜ = π’Œ=Β± 𝟏 𝟐 [AQA Worksheet] 𝑓 π‘₯ =2 π‘₯ 3 βˆ’250. Work out π‘₯ when 𝑓 π‘₯ =0 𝟐 𝒙 πŸ‘ βˆ’πŸπŸ“πŸŽ=𝟎 β†’ 𝒙=πŸ“ [AQA Worksheet] 𝑓 π‘₯ = π‘₯ 2 +π‘Žπ‘₯βˆ’8. If 𝑓 βˆ’3 =13, determine the value of π‘Ž. πŸ—βˆ’πŸ‘π’‚βˆ’πŸ–=πŸπŸ‘ 𝒂=βˆ’πŸ’ If 𝑓 π‘₯ =5π‘₯+2, determine the following, simplifying where possible. 𝑓 π‘₯+1 =πŸ“ 𝒙+𝟏 +𝟐=πŸ“π’™+πŸ• 𝑓 π‘₯ 2 =πŸ“ 𝒙 𝟐 +𝟐 [AQA IGCSEFM June 2012 Paper 2] 𝑓 π‘₯ =3π‘₯βˆ’5 for all values of π‘₯. Solve 𝑓 π‘₯ 2 =43 πŸ‘ 𝒙 𝟐 βˆ’πŸ“=πŸ’πŸ‘ 𝒙=Β±πŸ’ 1 4 ? ? ? ? 2 5 ? ? ? ? ? 6 3 ? ?

9 Exercise 1 ? ? ? ? ? ? ? ? (exercises on provided sheet)
[Edexcel Specimen Papers Set 1, Paper 2H Q18] 𝑓 π‘₯ =3 π‘₯ 2 βˆ’2π‘₯βˆ’8 Express 𝑓 π‘₯+2 in the form π‘Ž π‘₯ 2 +𝑏π‘₯ πŸ‘ 𝒙 𝟐 +πŸπŸŽπ’™ [Senior Kangaroo 2011 Q20] The polynomial 𝑓 π‘₯ is such that 𝑓 π‘₯ 2 +1 = π‘₯ 4 +4 π‘₯ 2 and 𝑓 π‘₯ 2 βˆ’1 =π‘Ž π‘₯ 4 +4𝑏 π‘₯ 2 +𝑐. What is the value of π‘Ž 2 + 𝑏 2 + 𝑐 2 ? 𝒇 𝒙 𝟐 +𝟏 = 𝒙 𝟐 ( 𝒙 𝟐 +πŸ’) By letting π’š= 𝒙 𝟐 +𝟏: 𝒇 π’š =(π’šβˆ’πŸ)(π’š+πŸ‘) Thus 𝒇 𝒙 𝟐 βˆ’πŸ =𝒇 π’šβˆ’πŸ = π’šβˆ’πŸ‘ π’š+𝟏 = 𝒙 𝟐 βˆ’πŸ 𝒙 𝟐 +𝟐 = 𝒙 πŸ’ βˆ’πŸ’ 𝒂=𝟏, 𝒃=𝟎, 𝒄=βˆ’πŸ’ 𝒂 𝟐 + 𝒃 𝟐 + 𝒄 𝟐 =πŸπŸ• 9 [AQA Worksheet] 𝑓 π‘₯ = π‘₯ 2 +3π‘₯βˆ’10 Show that 𝑓 π‘₯+2 =π‘₯ π‘₯+7 𝒇 𝒙+𝟐 = 𝒙+𝟐 𝟐 +πŸ‘ 𝒙+𝟐 βˆ’πŸπŸŽ = 𝒙 𝟐 +πŸ’π’™+πŸ’+πŸ‘π’™+πŸ”βˆ’πŸπŸŽ = 𝒙 𝟐 +πŸ•π’™=𝒙 𝒙+πŸ• If 𝑓 π‘₯ =2π‘₯βˆ’1 determine: (a) 𝑓 2π‘₯ =πŸ’π’™βˆ’πŸ (b) 𝑓 π‘₯ 2 =𝟐 𝒙 𝟐 βˆ’πŸ (c) 𝑓 2π‘₯βˆ’1 =𝟐 πŸπ’™βˆ’πŸ βˆ’πŸ=πŸ’π’™βˆ’πŸ‘ (d) 𝑓 1+2𝑓 π‘₯βˆ’1 =𝒇 πŸ’π’™βˆ’πŸ“ =πŸ–π’™βˆ’πŸπŸ (e) Solve 𝑓 π‘₯+1 +𝑓 π‘₯βˆ’1 =0 𝟐 𝒙+𝟏 βˆ’πŸ+𝟐 π’™βˆ’πŸ βˆ’πŸ=𝟎 πŸ’π’™βˆ’πŸ=𝟎 β†’ 𝒙= 𝟏 𝟐 7 ? ? N 8 ? ? ? ? ? ?

10 Inverse Functions A function takes and input and produces an output. The inverse of a function does the opposite: it describes how we get from the output back to the input. Input Output Γ—3 2 6 ? Γ·3 Bro-notation: The -1 notation means that we apply the function β€œ-1 times”, i.e. once backwards! You’ve actually seen this before, remember sin βˆ’1 (π‘₯) from trigonometry to mean β€œinverse sin”? It’s possible to have 𝑓 2 (π‘₯), we’ll see this when we cover composite functions. So if 𝑓 π‘₯ =3π‘₯, then the inverse function is : 𝒇 βˆ’πŸ 𝒙 = 𝒙 πŸ‘ ?

11 Quickfire Questions ? 𝑓 π‘₯ =π‘₯+5 𝑓 βˆ’1 π‘₯ =π’™βˆ’πŸ“ 𝑓 βˆ’1 π‘₯ = 𝒙+𝟏 πŸ‘ ? 𝑓 π‘₯ =3π‘₯βˆ’1
In your head, find the inverse functions, by thinking what the original functions does, and what the reverse process would therefore be. ? 𝑓 π‘₯ =π‘₯+5 𝑓 βˆ’1 π‘₯ =π’™βˆ’πŸ“ 𝑓 βˆ’1 π‘₯ = 𝒙+𝟏 πŸ‘ ? 𝑓 π‘₯ =3π‘₯βˆ’1 ? 𝑓 π‘₯ = π‘₯ +3 𝑓 βˆ’1 π‘₯ = π’™βˆ’πŸ‘ 𝟐 𝑓 βˆ’1 π‘₯ = 𝟏 𝒙 𝑓 π‘₯ = 1 π‘₯ ? Bro Fact: If a function is the same as its inverse, it is known as self-inverse. 𝑓 π‘₯ =1βˆ’π‘₯ is also a self-inverse function.

12 Full Method ? ? ? If 𝑓 π‘₯ = π‘₯ 5 +1, find 𝑓 βˆ’1 (π‘₯). 𝑦= π‘₯ 5 +1
STEP 1: Write the output 𝑓(π‘₯) as 𝑦 This is purely for convenience. ? π‘¦βˆ’1= π‘₯ 5 5π‘¦βˆ’5=π‘₯ STEP 2: Get the input in terms of the output (make π‘₯ the subject). This is because the inverse function is the reverse process, i.e. finding the input π‘₯ in terms of the output 𝑦. ? 𝑓 βˆ’1 π‘₯ =5π‘₯βˆ’5 STEP 3: Swap 𝑦 back for π‘₯ and π‘₯ back for 𝑓 βˆ’1 π‘₯ . This is because the input to a function is generally written as π‘₯ rather than 𝑦. But technically 𝑓 βˆ’1 𝑦 =5π‘¦βˆ’5 would be correct!

13 Harder One If 𝑓 π‘₯ = π‘₯+1 π‘₯βˆ’2 , find 𝑓 βˆ’1 (π‘₯). ? 𝑦= π‘₯+1 π‘₯βˆ’2 π‘₯π‘¦βˆ’2𝑦=π‘₯+1 π‘₯π‘¦βˆ’π‘₯=1+2𝑦 π‘₯ π‘¦βˆ’1 =1+2𝑦 π‘₯= 1+2𝑦 π‘¦βˆ’1 𝑓 βˆ’1 π‘₯ = 1+2π‘₯ π‘₯βˆ’1

14 Test Your Understanding
If 𝑓 π‘₯ = 2π‘₯+1 3 , find 𝑓 βˆ’1 (4). If 𝑓 π‘₯ = π‘₯ 2π‘₯βˆ’1 , find 𝑓 βˆ’1 (π‘₯). ? ? π’š= πŸπ’™+𝟏 πŸ‘ πŸ‘π’š=πŸπ’™+𝟏 πŸ‘π’šβˆ’πŸ=πŸπ’™ 𝒙= πŸ‘π’šβˆ’πŸ 𝟐 𝒇 βˆ’πŸ 𝒙 = πŸ‘π’™βˆ’πŸ 𝟐 𝒇 βˆ’πŸ πŸ‘ = 𝟏𝟏 𝟐 π’š= 𝒙 πŸπ’™βˆ’πŸ πŸπ’™π’šβˆ’π’š=𝒙 πŸπ’™π’šβˆ’π’™=π’š 𝒙 πŸπ’šβˆ’πŸ =π’š 𝒙= π’š πŸπ’šβˆ’πŸ 𝒇 βˆ’πŸ 𝒙 = 𝒙 πŸπ’™βˆ’πŸ

15 Exercise 2 ? ? ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet)
Find 𝑓 βˆ’1 (π‘₯) for the following functions. 𝑓 π‘₯ =5π‘₯ 𝒇 βˆ’πŸ 𝒙 = 𝒙 πŸ“ 𝑓 π‘₯ =1+π‘₯ 𝒇 βˆ’πŸ 𝒙 =π’™βˆ’πŸ 𝑓 π‘₯ =6π‘₯βˆ’ 𝒇 βˆ’πŸ 𝒙 = 𝒙+πŸ’ πŸ” 𝑓 π‘₯ = π‘₯ 𝒇 βˆ’πŸ 𝒙 =πŸ‘π’™βˆ’πŸ• 𝑓 π‘₯ =5 π‘₯ 𝒇 βˆ’πŸ 𝒙 = π’™βˆ’πŸ πŸ“ 𝟐 𝑓 π‘₯ =10βˆ’3π‘₯ 𝒇 βˆ’πŸ 𝒙 = πŸπŸŽβˆ’π’™ πŸ‘ [Edexcel IGCSE Jan2016(R)-3H Q16c] 𝑓 π‘₯ = 2π‘₯ π‘₯βˆ’1 Find 𝑓 βˆ’1 π‘₯ = 𝒙 π’™βˆ’πŸ Find 𝑓 βˆ’1 (π‘₯) for the following functions. 𝑓 π‘₯ = π‘₯ π‘₯ 𝒇 βˆ’πŸ 𝒙 = πŸ‘π’™ πŸβˆ’π’™ 𝑓 π‘₯ = π‘₯βˆ’2 π‘₯ 𝒇 βˆ’πŸ 𝒙 = 𝟐 πŸβˆ’π’™ 𝑓 π‘₯ = 2π‘₯βˆ’1 π‘₯βˆ’ 𝒇 βˆ’πŸ 𝒙 = π’™βˆ’πŸ π’™βˆ’πŸ 𝑓 π‘₯ = 1βˆ’π‘₯ 3π‘₯ 𝒇 βˆ’πŸ 𝒙 = πŸβˆ’π’™ πŸ‘π’™+𝟏 𝑓 π‘₯ = 3π‘₯ 3+2π‘₯ 𝒇 βˆ’πŸ 𝒙 = πŸπ’™ πŸ‘βˆ’πŸ‘π’™ Find the value of π‘Ž for which 𝑓 π‘₯ = π‘₯ π‘₯+π‘Ž is a self inverse function. 𝒇 βˆ’πŸ 𝒙 = 𝒂𝒙 πŸβˆ’π’™ If self-inverse: 𝒙 𝒙+𝒂 ≑ 𝒂𝒙 πŸβˆ’π’™ 𝒂 𝒙 𝟐 + 𝒂 𝟐 π’™β‰‘π’™βˆ’ 𝒙 𝟐 For 𝒙 𝟐 and 𝒙 terms to match, 𝒂=βˆ’πŸ. 1 3 ? ? a a b ? ? b c ? ? d ? c ? e ? d f ? ? e 2 N ? ?

16 𝑓𝑔 2 = 𝑓 π‘₯ =3π‘₯+1 𝑔 π‘₯ = π‘₯ 2 Composite Functions 49? 13?
Have a guess! (Click your answer) 𝑓𝑔 2 = 49? 13? 𝑓𝑔(2) means 𝑓 𝑔 2 , i.e. β€œπ‘“ of 𝑔 of 2”. We therefore apply the functions to the input in sequence from right to left.

17 𝑓 π‘₯ =3π‘₯+1 𝑔 π‘₯ = π‘₯ 2 Examples ? ? ? ? 𝑓𝑔 5 =𝒇 π’ˆ πŸ“ =𝒇 πŸπŸ“ =πŸ•πŸ”
Bro Tip: I highly encourage you to write this first. It will help you when you come to the algebraic ones. Determine: 𝑓𝑔 5 =𝒇 π’ˆ πŸ“ =𝒇 πŸπŸ“ =πŸ•πŸ” 𝑔𝑓 βˆ’1 =π’ˆ 𝒇 βˆ’πŸ =π’ˆ βˆ’πŸ =πŸ’ 𝑓𝑓 4 =𝒇 𝒇 πŸ’ =𝒇 πŸπŸ‘ =πŸ’πŸŽ 𝑔𝑓 π‘₯ =π’ˆ 𝒇 𝒙 =π’ˆ πŸ‘π’™+𝟏 = πŸ‘π’™+𝟏 𝟐 ? ? ? Bro Note: This can also be written as 𝑓 2 (π‘₯), but you won’t encounter this notation in GCSE/IGCSE FM. ?

18 𝑓 π‘₯ =2π‘₯+1 𝑔 π‘₯ = 1 π‘₯ More Algebraic Examples ? ? ? ? ?
Determine: 𝑓𝑔 π‘₯ =𝒇 π’ˆ 𝒙 =𝒇 𝟏 𝒙 =𝟐 𝟏 𝒙 +𝟏= 𝟐 𝒙 +𝟏 𝑔𝑓 π‘₯ =π’ˆ πŸπ’™+𝟏 = 𝟏 πŸπ’™+𝟏 𝑓𝑓 π‘₯ =𝒇 πŸπ’™+𝟏 =𝟐 πŸπ’™+𝟏 +𝟏=πŸ’π’™+πŸ‘ 𝑔𝑔 π‘₯ =π’ˆ 𝟏 𝒙 = 𝟏 𝟏 𝒙 =𝒙 ? ? ? ? ?

19 Test Your Understanding
If 𝑓 π‘₯ = 2 π‘₯+1 and 𝑔 π‘₯ = π‘₯ 2 βˆ’1, determine 𝑓𝑔(π‘₯). 𝒇 𝒙 𝟐 βˆ’πŸ = 𝟐 𝒙 𝟐 βˆ’πŸ+𝟏 = 𝟐 𝒙 𝟐 1 2 ? 3 A function 𝑓 is such that 𝑓 π‘₯ =3π‘₯+1 The function 𝑔 is such that 𝑔 π‘₯ =π‘˜ π‘₯ 2 where π‘˜ is a constant. Given that 𝑓𝑔 3 =55, determine the value of π‘˜. π’‡π’ˆ πŸ‘ =𝒇 πŸ—π’Œ =πŸ‘ πŸ—π’Œ +𝟏 =πŸπŸ•π’Œ+𝟏=πŸ“πŸ“ π’Œ=𝟐 ? π’‡π’ˆ βˆ’πŸ‘ =𝒇 π’ˆ βˆ’πŸ‘ =𝒇 βˆ’πŸ =πŸ— ?

20 Exercise 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet) 1
If 𝑓 π‘₯ =3π‘₯ and 𝑔 π‘₯ =π‘₯+1, determine: 𝑓𝑔 2 =πŸ— 𝑔𝑓 4 =πŸπŸ‘ 𝑓𝑔 π‘₯ =πŸ‘π’™+πŸ‘ 𝑔𝑓 π‘₯ =πŸ‘π’™+𝟏 𝑔𝑔 π‘₯ =𝒙+𝟐 If 𝑓 π‘₯ =2π‘₯+1 and 𝑔 π‘₯ =3π‘₯+1 determine: 𝑓𝑔 π‘₯ =πŸ”π’™+πŸ‘ 𝑔𝑓 π‘₯ =πŸ”π’™+πŸ’ 𝑓𝑓 π‘₯ =πŸ’π’™+πŸ‘ If 𝑓 π‘₯ = π‘₯ 2 βˆ’2π‘₯ and 𝑔 π‘₯ =π‘₯+1, find 𝑓𝑔(π‘₯), simplifying your expression. 𝒇 𝒙+𝟏 = 𝒙+𝟏 𝟐 βˆ’πŸ 𝒙+𝟏 = 𝒙 𝟐 +πŸπ’™+πŸβˆ’πŸπ’™βˆ’πŸ = 𝒙 𝟐 βˆ’πŸ If 𝑓 π‘₯ =π‘₯+π‘˜ and 𝑔 π‘₯ = π‘₯ 2 and 𝑔𝑓 3 =16, find the possible values of π‘˜. π’ˆπ’‡ πŸ‘ =π’ˆ πŸ‘+π’Œ = πŸ‘+π’Œ 𝟐 =πŸπŸ” π’Œ=𝟏,βˆ’πŸ• If 𝑓 π‘₯ =2(π‘₯+π‘˜) and 𝑔 π‘₯ = π‘₯ 2 βˆ’π‘₯ and 𝑓𝑔 3 =30, find π‘˜. 𝒇 π’ˆ πŸ‘ =𝒇 πŸ” =𝟐 πŸ”+π’Œ =πŸ‘πŸŽ π’Œ=πŸ— 6 Let 𝑓 π‘₯ =π‘₯+1 and 𝑔 π‘₯ = π‘₯ 2 +1. If 𝑔𝑓 π‘₯ =17, determine the possible values of π‘₯. π’ˆπ’‡ 𝒙 = 𝒙+𝟏 𝟐 +𝟏=πŸπŸ• 𝒙 𝟐 +πŸπ’™+𝟐=πŸπŸ• 𝒙 𝟐 +πŸπ’™βˆ’πŸπŸ“=𝟎 𝒙+πŸ“ π’™βˆ’πŸ‘ =𝟎 𝒙=βˆ’πŸ“ 𝒐𝒓 𝒙=πŸ‘ Let 𝑓 π‘₯ = π‘₯ 2 +3π‘₯ and 𝑔 π‘₯ =π‘₯βˆ’2. If 𝑓𝑔 π‘₯ =0, determine the possible values of π‘₯. 𝒙=βˆ’πŸ 𝒐𝒓 𝒙=𝟐 [Based on MAT question] 𝑓 π‘₯ =π‘₯+1 and 𝑔 π‘₯ =2π‘₯ Let 𝑓 𝑛 (π‘₯) means that you apply the function 𝑓 𝑛 times. a) Find 𝑓 𝑛 (π‘₯) in terms of π‘₯ and 𝑛. =𝒙+𝒏 b) Note that 𝑔 𝑓 2 𝑔 π‘₯ =4π‘₯+4. Find all other ways of combining 𝑓 and 𝑔 that result in the function 4π‘₯+4. π’ˆ 𝟐 𝒇, 𝒇 𝟐 π’ˆπ’‡π’ˆ, 𝒇 πŸ’ π’ˆ 𝟐 ? ? ? ? ? ? 2 ? ? ? 7 3 ? ? N 4 ? 5 ? ?

21 This be ye end of GCSE functions content.
Beyond this point there be IGCSE Further Maths. Yarr.

22 #4 :: Piecewise Functions
Sometimes functions are defined in β€˜pieces’, with a different function for different ranges of π‘₯ values. Sketch > Sketch > Sketch > (2, 9) (0, 5) (-1, 0) (5, 0)

23 Test Your Understanding
𝑓 π‘₯ = π‘₯ 2 0≀π‘₯<1 1 1≀π‘₯<2 3βˆ’π‘₯ 2≀π‘₯<3 Sketch Sketch Sketch This example was used on the specification itself! (2, 1) (1, 1) (3, 0)

24 Exercise 4 b ? c ? ? a ? (Exercises on provided sheet)
[Jan 2013 Paper 2] A function 𝑓(π‘₯) is defined as: 𝑓 π‘₯ = 4 π‘₯<βˆ’2 π‘₯ 2 βˆ’2≀π‘₯≀2 12βˆ’4π‘₯ π‘₯>2 Draw the graph of 𝑦=𝑓(π‘₯) for βˆ’4≀π‘₯≀4 Use your graph to write down how many solutions there are to 𝑓 π‘₯ = sols Solve 𝑓 π‘₯ =βˆ’ πŸπŸβˆ’πŸ’π’™=βˆ’πŸπŸŽ →𝒙= 𝟏𝟏 𝟐 [June 2013 Paper 2] A function 𝑓(π‘₯) is defined as: 𝑓 π‘₯ = π‘₯+3 βˆ’3≀π‘₯<0 3 0≀π‘₯<1 5βˆ’2π‘₯ 1≀π‘₯≀2 Draw the graph of 𝑦=𝑓(π‘₯) for βˆ’3≀π‘₯<2 1 2 b ? ? c ? a ?

25 Exercise 4 ? Sketch ? ? (Exercises on provided sheet)
[Specimen 1 Q4] A function 𝑓(π‘₯) is defined as: 𝑓 π‘₯ = 3π‘₯ 0≀π‘₯<1 3 1≀π‘₯<3 12βˆ’3π‘₯ 3≀π‘₯≀4 Calculate the area enclosed by the graph of 𝑦=𝑓 π‘₯ and the π‘₯βˆ’axis. 3 [Set 1 Paper 1] A function 𝑓(π‘₯) is defined as: 𝑓 π‘₯ = 3 0≀π‘₯<2 π‘₯+1 2≀π‘₯<4 9βˆ’π‘₯ 4≀π‘₯≀9 Draw the graph of 𝑦=𝑓(π‘₯) for 0≀π‘₯≀9. 4 ? Sketch ? Area = πŸ— ?

26 Exercise 4 ? ? (Exercises on provided sheet) [AQA Worksheet Q9] 6
𝑓 π‘₯ = βˆ’ π‘₯ 2 0≀π‘₯<2 βˆ’4 2≀π‘₯<3 2π‘₯βˆ’10 3≀π‘₯≀5 Draw the graph of 𝑓(π‘₯) from 0≀π‘₯≀5. 6 [AQA Worksheet Q10] 𝑓 π‘₯ = 2π‘₯ 0≀π‘₯<1 3βˆ’π‘₯ 1≀π‘₯<4 π‘₯βˆ’7 3 4≀π‘₯≀7 5 ? 2 -1 -2 -3 -4 3 7 -1 Show that π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝐴:π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝐡= 3:2 Area of 𝑨= 𝟏 𝟐 Γ—πŸ‘Γ—πŸ=πŸ‘ Area of 𝑩= 𝟏 𝟐 Γ—πŸ’Γ—πŸ=𝟐 ?

27 Domain and Range 1 -1 𝑓 π‘₯ = π‘₯ 2 2.89 1.7 4 2 3.1 9.61 ... ... Inputs
Outputs 𝑓 π‘₯ = π‘₯ 2 2.89 1.7 4 2 3.1 9.61 ... ... ! The domain of a function is the set of possible inputs. ! The range of a function is the set of possible outputs.

28 Example 𝑓 π‘₯ = π‘₯ 2 Sketch: for all π‘₯ Suitable Domain: Range: 𝑓 π‘₯ β‰₯0 ? ?
𝑦 𝑓 π‘₯ = π‘₯ 2 Sketch: π‘₯ Bro Note: By β€˜suitable’, I mean the largest possible set of values that could be input into the function. Suitable Domain: for all π‘₯ ? We can use any real number as the input! In β€˜proper’ maths we’d use π‘₯βˆˆβ„ to mean β€œπ‘₯ can be any element in the set of real numbers”, but the syllabus is looking for β€œfor all π‘₯”. Range: ? 𝑓 π‘₯ β‰₯0 Look at the 𝑦 values on the graph. The output has to be positive, since it’s been squared. B Bro Tip: Note that the domain is in terms of π‘₯ and the range in terms of 𝑓 π‘₯ .

29 Test Your Understanding
? 𝑦 Sketch: 𝑓 π‘₯ = π‘₯ π‘₯ Suitable Domain: π‘₯β‰₯0 ? Presuming the output has to be a real number, we can’t input negative numbers into our function. Range: ? 𝑓 π‘₯ β‰₯0 The output, again, can only be positive.

30 Mini-Exercise In pairs, work out a suitable domain and the range of each function. A sketch may help with each one. 3 1 2 Function 𝑓 π‘₯ = 1 π‘₯ Domain For all π‘₯ except 0 Range For all 𝑓 π‘₯ except 0 Function 𝑓 π‘₯ =2π‘₯ Domain For all π‘₯ Range For all 𝑓(π‘₯) Function 𝑓 π‘₯ = 2 π‘₯ Domain For all π‘₯ Range 𝑓 π‘₯ >0 ? ? ? 4 5 6 Function 𝑓 π‘₯ = sin π‘₯ Domain For all π‘₯ Range βˆ’1≀𝑓 π‘₯ ≀1 Function 𝑓 π‘₯ =2 cos π‘₯ Domain For all π‘₯ Range βˆ’2≀𝑓 π‘₯ ≀2 Function 𝑓 π‘₯ = π‘₯ 3 +1 Domain For all π‘₯ Range For all 𝑓(π‘₯) ? ? ? 7 8 Function 𝑓 π‘₯ = 1 π‘₯βˆ’2 +1 Domain For all π‘₯ except 2 Range For all 𝑓 π‘₯ except 1 Function 𝑓 π‘₯ = 2cos π‘₯+1 Domain π‘₯>βˆ’1 Range βˆ’2≀𝑓 π‘₯ ≀2 ? ?

31 Range of Quadratics A common exam question is to determine the range of a quadratic. The sketch shows the function 𝑦=𝑓(π‘₯) where 𝑓 π‘₯ = π‘₯ 2 βˆ’4π‘₯+7. Determine the range of 𝑓(π‘₯). 𝑦 ? We need the minimum point, since from the graph we can see that π’š (i.e. 𝒇(𝒙)) can be anything greater than this. 𝒇 𝒙 = π’™βˆ’πŸ 𝟐 +πŸ‘ The minimum point is (𝟐,πŸ‘) thus the range is: 𝒇 𝒙 β‰₯πŸ‘ (note the β‰₯ rather than >) 3 π‘₯ An alternative way of thinking about it, once you’ve completed the square, is that anything squared is at least 0. So if π‘₯βˆ’2 3 is at least 0, then clearly π‘₯βˆ’ is at least 3.

32 Test Your Understanding
𝑦 The sketch shows the function 𝑦=𝑓(π‘₯) where 𝑓 π‘₯ = π‘₯+2 π‘₯βˆ’4 . Determine the range of 𝑓(π‘₯). ? π‘₯+2 π‘₯βˆ’4 = π‘₯ 2 βˆ’2π‘₯βˆ’ = π‘₯βˆ’1 2 βˆ’9 Therefore 𝑓 π‘₯ β‰₯βˆ’9 π‘₯ 1,βˆ’9 𝑦 The sketch shows the function 𝑦=𝑓(π‘₯) where 𝑓 π‘₯ =21+4π‘₯βˆ’ π‘₯ 2 . Determine the range of 𝑓(π‘₯). 2,25 ? βˆ’ π‘₯ 2 βˆ’4π‘₯βˆ’21 =βˆ’ π‘₯βˆ’2 2 βˆ’4βˆ’21 =25βˆ’ π‘₯βˆ’2 2 Therefore 𝑓 π‘₯ ≀25 π‘₯

33 Range for Restricted Domains
Some questions are a bit jammy by restricting the domain. Look out for this, because it affects the domain! 𝑓 π‘₯ = π‘₯ 2 +4π‘₯+3, π‘₯β‰₯1 Determine the range of 𝑓(π‘₯). ? 𝑦 Notice how the domain is 𝒙β‰₯𝟏. 𝒇 𝒙 =(𝒙+𝟏)(𝒙+πŸ‘) When 𝒙=𝟏, π’š= 𝟏 𝟐 +πŸ’+πŸ‘=πŸ– Sketching the graph, we see that when 𝒙=𝟏, the function is increasing. Therefore when 𝒙β‰₯𝟏, 𝒇 𝒙 β‰₯πŸ– π‘₯ βˆ’3 βˆ’1 1

34 Test Your Understanding
𝑓 π‘₯ = π‘₯ 2 βˆ’3, π‘₯β‰€βˆ’2 Determine the range of 𝑓(π‘₯). 𝑓 π‘₯ =3π‘₯βˆ’2, 0≀π‘₯<4 Determine the range of 𝑓(π‘₯). ? ? 𝑦 When π‘₯=0, 𝑓 π‘₯ =βˆ’2 When π‘₯=4, 𝑓 π‘₯ =10 Range: βˆ’πŸβ‰€π’‡ 𝒙 <𝟏𝟎 π‘₯ βˆ’2 When π‘₯=βˆ’2, 𝑓 π‘₯ =1 As π‘₯ decreases from -2, 𝑓(π‘₯) is increasing. Therefore: 𝑓 π‘₯ β‰₯1

35 Range of Trigonometric Functions
90Β° 180Β° 270Β° 360Β° Suppose we restricted the domain in different ways. Determine the range in each case (or vice versa). Ignore angles below 0 or above 360. Domain Range For all π‘₯ (i.e. unrestricted) βˆ’1≀𝑓 π‘₯ ≀1 180≀π‘₯≀360 βˆ’1≀𝑓 π‘₯ ≀0 0≀π‘₯≀180 0≀𝑓 π‘₯ ≀1 ? ? ?

36 Range of Piecewise Functions
It’s a simple case of just sketching the full function. The sketch shows the graph of 𝑦=𝑓(π‘₯) with the domain 0≀π‘₯≀9 𝑓 π‘₯ = 3 0≀π‘₯<2 π‘₯+1 2≀π‘₯<4 9βˆ’π‘₯ 4≀π‘₯≀9 Determine the range of 𝑓(π‘₯). Graph ? Range ? Range: πŸŽβ‰€π’‡ 𝒙 β‰€πŸ“

37 Test Your Understanding
The function 𝑓(π‘₯) is defined for all π‘₯: 𝑓 π‘₯ = 4 π‘₯<βˆ’2 π‘₯ 2 βˆ’2≀π‘₯≀2 12βˆ’4π‘₯ π‘₯>2 Determine the range of 𝑓(π‘₯). Graph ? Range ? Range: 𝒇 𝒙 β‰€πŸ’

38 Exercise 5 ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet)
Work out the range for each of these functions. (a) 𝑓 π‘₯ = π‘₯ for all π‘₯ 𝒇 𝒙 β‰₯πŸ” (b) 𝑓 π‘₯ =3π‘₯βˆ’5, βˆ’2≀π‘₯≀6 βˆ’πŸπŸβ‰€π’‡ 𝒙 β‰€πŸπŸ‘ (c) 𝑓 π‘₯ =3 π‘₯ 4 , π‘₯<βˆ’2 𝒇 𝒙 >πŸ’πŸ– (a) 𝑓 π‘₯ = π‘₯+2 π‘₯βˆ’3 Give a reason why π‘₯>0 is not a suitable domain for 𝑓(π‘₯). It would include 3, for which 𝒇(𝒙) is undefined. (b) Give a possible domain for 𝑓 π‘₯ = π‘₯βˆ’ 𝒙β‰₯πŸ“ 𝑓 π‘₯ =3βˆ’2π‘₯, π‘Ž<π‘₯<𝑏 The range of 𝑓(π‘₯) is βˆ’5<𝑓 π‘₯ <5 Work out π‘Ž and 𝑏. 𝒂=βˆ’πŸ, 𝒃=πŸ’ 4 [Set 1 Paper 2] (a) The function 𝑓(π‘₯) is defined as: 𝑓 π‘₯ =22βˆ’7π‘₯, βˆ’2≀π‘₯≀𝑝 The range of 𝑓(π‘₯) is βˆ’13≀𝑓 π‘₯ ≀36 Work out the value of 𝑝. 𝒑=πŸ“ (b) The function 𝑔(π‘₯) is defined as 𝑔 π‘₯ = π‘₯ 2 βˆ’4π‘₯+5 for all π‘₯. (i) Express 𝑔(π‘₯) in the form π‘₯βˆ’π‘Ž 2 +𝑏 π’ˆ 𝒙 = π’™βˆ’πŸ 𝟐 +𝟏 (ii) Hence write down the range of 𝑔(π‘₯). π’ˆ 𝒙 β‰₯𝟏 [June 2012 Paper 1] 𝑓 π‘₯ =2 π‘₯ 2 +7 for all values of π‘₯. (a) What is the value of 𝑓 βˆ’1 ? 𝒇 βˆ’πŸ =πŸ— (b) What is the range of 𝑓(π‘₯)? 𝒇 𝒙 β‰₯πŸ• 1 ? ? ? ? 2 ? ? ? 5 ? ? 3 ? ?

39 Exercise 5 ? ? ? ? ? (exercises on provided sheet)
[Jan 2013 Paper 2] 𝑓 π‘₯ = sin π‘₯ °≀π‘₯≀360Β° 𝑔 π‘₯ = cos π‘₯ °≀π‘₯β‰€πœƒ (a) What is the range of 𝑓(π‘₯)? βˆ’πŸβ‰€π’‡ 𝒙 β‰€πŸŽ (b) You are given that 0≀𝑔 π‘₯ ≀1. Work out the value of πœƒ. 𝜽=πŸ—πŸŽΒ° By completing the square or otherwise, determine the range of the following functions: (a) 𝑓 π‘₯ = π‘₯ 2 βˆ’2π‘₯+5, for all π‘₯ = π’™βˆ’πŸ 𝟐 +πŸ’ Range: 𝒇 𝒙 β‰₯πŸ’ (b) 𝑓 π‘₯ = π‘₯ 2 +6π‘₯βˆ’2, for all π‘₯ = 𝒙+πŸ‘ 𝟐 βˆ’πŸπŸ Range: 𝒇 𝒙 β‰₯βˆ’πŸπŸ 6 8 ? ? Here is a sketch of 𝑓 π‘₯ = π‘₯ 2 +6π‘₯+π‘Ž for all π‘₯, where π‘Ž is a constant. The range of 𝑓(π‘₯) is 𝑓 π‘₯ β‰₯11. Work out the value of π‘Ž. 𝒇 𝒙 = 𝒙+πŸ‘ 𝟐 βˆ’πŸ—+𝒂 βˆ’πŸ—+𝒂=𝟏𝟏 𝒂=𝟐𝟎 7 ? ? ?

40 Exercise 5 ? ? ? ? ? (exercises on provided sheet) 9 10
The straight line shows a sketch of 𝑦=𝑓(π‘₯) for the full domain of the function. (a) State the domain of the function. πŸβ‰€π’‡ 𝒙 β‰€πŸπŸ’ (b) Work out the equation of the line. 𝒇 𝒙 =βˆ’πŸπ’™+𝟏𝟎 𝑓(π‘₯) is a quadratic function with domain all real values of π‘₯. Part of the graph of 𝑦=𝑓 π‘₯ is shown. (a) Write down the range of 𝑓(π‘₯). 𝒇 𝒙 β‰€πŸ’ (b) Use the graph to find solutions of the equation 𝑓 π‘₯ =1. 𝒙=βˆ’πŸŽ.πŸ•, 𝟐.πŸ• (c) Use the graph to solve 𝑓 π‘₯ <0. 𝒙<βˆ’πŸ 𝒐𝒓 𝒙>πŸ‘ ? ? ? ? ?

41 Exercise 5 ? ? ? (exercises on provided sheet)
The function 𝑓(π‘₯) is defined as: 𝑓 π‘₯ = π‘₯ 2 βˆ’4 0≀π‘₯<3 14βˆ’3π‘₯ 3≀π‘₯≀5 Work out the range of 𝑓 π‘₯ . 𝒇(𝒙)β‰€πŸ“ The function 𝑓(π‘₯) has the domain βˆ’3≀π‘₯≀3 and is defined as: 𝑓 π‘₯ = π‘₯ 2 +3π‘₯+2 βˆ’3≀π‘₯<0 2+π‘₯ 0≀π‘₯≀3 Work out the range of 𝑓 π‘₯ . βˆ’ 𝟏 πŸ’ ≀𝒇 𝒙 β‰€πŸ“ 11 13 ? 12 [June 2012 Paper 2] A sketch of 𝑦=𝑔(π‘₯) for domain 0≀π‘₯≀8 is shown. The graph is symmetrical about π‘₯=4. The range of 𝑔(π‘₯) is 0≀𝑔 π‘₯ ≀12. Work out the function 𝑔(π‘₯). 𝑔 π‘₯ = ? 0≀π‘₯≀4 ? 4<π‘₯≀8 π’ˆ 𝒙 = πŸ‘π’™ πŸŽβ‰€π’™β‰€πŸ’ πŸπŸ’βˆ’πŸ‘π’™ πŸ’<π’™β‰€πŸ– ? ?

42 Constructing a function from a domain/range
June 2013 Paper 2 What would be the simplest function to use that has this domain/range? A straight line! Note, that could either be going up or down (provided it starts and ends at a corner) What is the equation of this? π’Ž= πŸ– πŸ’ =𝟐 π’šβˆ’πŸ‘=𝟐 π’™βˆ’πŸ π’š=πŸπ’™+𝟏 𝒇 𝒙 =πŸπ’™+𝟏 𝑦 ? 11 ? 3 π‘₯ 1 5

43 Constructing a function from a domain/range
Sometimes there’s the additional constraint that the function is β€˜increasing’ or β€˜decreasing’. We’ll cover this in more depth when we do calculus, but the meaning of these words should be obvious. 𝑓 π‘₯ is a decreasing function with domain 4≀π‘₯≀6 and range 7≀𝑓 π‘₯ ≀19. ? 𝑦 π’Ž=βˆ’ 𝟏𝟐 𝟐 =βˆ’πŸ” π’šβˆ’πŸ•=βˆ’πŸ” π’™βˆ’πŸ” π’š=βˆ’πŸ”π’™+πŸ’πŸ‘ 𝒇 𝒙 =πŸ’πŸ‘βˆ’πŸ”π’™ 19 7 π‘₯ 4 6

44 Exercise 6 (exercises on provided sheet) 1 Domain is 1≀π‘₯<3. Range 1≀𝑓 π‘₯ ≀3. 𝑓(π‘₯) is an increasing function. 𝒇 𝒙 =𝒙 Domain is 1≀π‘₯≀3. Range 1≀𝑓 π‘₯ ≀3. 𝑓(π‘₯) is a decreasing function. 𝒇 𝒙 =πŸπŸ’βˆ’π’™ Domain is 5≀π‘₯≀7. Range 7≀𝑓 π‘₯ ≀11. 𝑓(π‘₯) is an increasing function. 𝒇 𝒙 =πŸπŸ’π’™βˆ’πŸ‘ Domain is 5≀π‘₯≀7. Range 7≀𝑓 π‘₯ ≀11. 𝑓(π‘₯) is a decreasing function. 𝒇 𝒙 =πŸπŸβˆ’πŸπ’™ Domain is βˆ’4≀π‘₯≀7. Range 4≀𝑓 π‘₯ ≀8. 𝑓(π‘₯) is a decreasing function. 𝒇 𝒙 = πŸ•πŸ 𝟏𝟏 βˆ’ πŸ’ 𝟏𝟏 𝒙 ? 2 ? 3 ? 4 ? 5 ?


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