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GCSE/IGCSE-FM Functions
Dr J Frost Last modified: 18th June 2017
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OVERVIEW #1: Understanding of functions #2: Inverse Functions GCSE
IGCSEFM GCSE #3: Composite Functions GCSE
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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
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Name of the function (usually π or π)
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 βnumber machineβ. Input Output f π₯ 2π₯ ? Input Output Name of the function (usually π or π) π(π₯)=2π₯
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π(π₯)=π₯2+2 Check Your Understanding ? ? ? ? What does this function do?
It squares the input then adds 2 to it. Q1 ? What is π(3)? π π = π π +π=ππ What is π(β5)? π βπ = βπ π +π=ππ If π π =38, what is π? π π +π=ππ So π=Β±π Q2 ? Q3 ? This question is asking the opposite, i.e. βwhat input π would give an output of 38?β Q4 ?
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Algebraic Inputs If π π₯ =π₯+1 what is: If π π₯ = π₯ 2 β1 what is: ? ? ? ?
If you change the input of the function (π₯), just replace each occurrence of π₯ in the output. If π π₯ =π₯+1 what is: If π π₯ = π₯ 2 β1 what is: π π₯β1 = πβπ +π=π π π₯ 2 = π π +π π π₯ 2 = π+π π π 2π₯ =ππ+π ? π π₯β1 = πβπ π βπ = π π βππ π 2π₯ = ππ π βπ =π π π βπ π π₯ 2 +1 = π π +π π βπ = π π +π π π ? ? ? ? ? ? If π π₯ =2π₯ what is: π π₯β1 =π πβπ =ππβπ π π₯ 2 =π π π π π₯ 2 = ππ π =π π π ? ? ?
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Test Your Understanding
If π π₯ =3π₯β1, determine: π π₯β1 =π πβπ βπ=ππβπ π 2π₯ =π ππ βπ=ππβπ π π₯ 3 =π π π βπ ? ? ? B If π π₯ =2π₯+1, solve π π₯ 2 =51 2 π₯ 2 +1=51 π₯=Β±5 ?
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Exercise 1 ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet)
If π π₯ =2π₯+5, find: π 3 =ππ π β1 =π π =π If π π₯ = π₯ 2 +5, find π β1 =π the possible values of π such that π π = π=Β±π The possible values of π such that π π = π=Β± π π [AQA Worksheet] π π₯ =2 π₯ 3 β250. Work out π₯ when π π₯ =0 π π π βπππ=π β π=π [AQA Worksheet] π π₯ = π₯ 2 +ππ₯β8. If π β3 =13, determine the value of π. πβππβπ=ππ π=βπ If π π₯ =5π₯+2, determine the following, simplifying where possible. π π₯+1 =π π+π +π=ππ+π π π₯ 2 =π π π +π [AQA IGCSEFM June 2012 Paper 2] π π₯ =3π₯β5 for all values of π₯. Solve π π₯ 2 =43 π π π βπ=ππ π=Β±π 1 4 ? ? ? ? 2 5 ? ? ? ? ? 6 3 ? ?
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Exercise 1 ? ? ? ? ? ? ? ? (exercises on provided sheet)
[Edexcel Specimen Papers Set 1, Paper 2H Q18] π π₯ =3 π₯ 2 β2π₯β8 Express π π₯+2 in the form π π₯ 2 +ππ₯ π π π +πππ [Senior Kangaroo 2011 Q20] The polynomial π π₯ is such that π π₯ 2 +1 = π₯ 4 +4 π₯ 2 and π π₯ 2 β1 =π π₯ 4 +4π π₯ 2 +π. What is the value of π 2 + π 2 + π 2 ? π π π +π = π π ( π π +π) By letting π= π π +π: π π =(πβπ)(π+π) Thus π π π βπ =π πβπ = πβπ π+π = π π βπ π π +π = π π βπ π=π, π=π, π=βπ π π + π π + π π =ππ 9 [AQA Worksheet] π π₯ = π₯ 2 +3π₯β10 Show that π π₯+2 =π₯ π₯+7 π π+π = π+π π +π π+π βππ = π π +ππ+π+ππ+πβππ = π π +ππ=π π+π If π π₯ =2π₯β1 determine: (a) π 2π₯ =ππβπ (b) π π₯ 2 =π π π βπ (c) π 2π₯β1 =π ππβπ βπ=ππβπ (d) π 1+2π π₯β1 =π ππβπ =ππβππ (e) Solve π π₯+1 +π π₯β1 =0 π π+π βπ+π πβπ βπ=π ππβπ=π β π= π π 7 ? ? N 8 ? ? ? ? ? ?
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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 Γ3 2 6 ? Γ·3 Bro-notation: The -1 notation means that we apply the function β-1 timesβ, i.e. once backwards! Youβve actually seen this before, remember sin β1 (π₯) from trigonometry to mean βinverse sinβ? Itβs possible to have π 2 (π₯), weβll see this when we cover composite functions. So if π π₯ =3π₯, then the inverse function is : π βπ π = π π ?
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Quickfire Questions ? π π₯ =π₯+5 π β1 π₯ =πβπ π β1 π₯ = π+π π ? π π₯ =3π₯β1
In your head, find the inverse functions, by thinking what the original functions does, and what the reverse process would therefore be. ? π π₯ =π₯+5 π β1 π₯ =πβπ π β1 π₯ = π+π π ? π π₯ =3π₯β1 ? π π₯ = π₯ +3 π β1 π₯ = πβπ π π β1 π₯ = π π π π₯ = 1 π₯ ? Bro Fact: If a function is the same as its inverse, it is known as self-inverse. π π₯ =1βπ₯ is also a self-inverse function.
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Full Method ? ? ? If π π₯ = π₯ 5 +1, find π β1 (π₯). π¦= π₯ 5 +1
STEP 1: Write the output π(π₯) as π¦ This is purely for convenience. ? π¦β1= π₯ 5 5π¦β5=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 π¦. ? π β1 π₯ =5π₯β5 STEP 3: Swap π¦ back for π₯ and π₯ back for π β1 π₯ . This is because the input to a function is generally written as π₯ rather than π¦. But technically π β1 π¦ =5π¦β5 would be correct!
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Harder One If π π₯ = π₯+1 π₯β2 , find π β1 (π₯). ? π¦= π₯+1 π₯β2 π₯π¦β2π¦=π₯+1 π₯π¦βπ₯=1+2π¦ π₯ π¦β1 =1+2π¦ π₯= 1+2π¦ π¦β1 π β1 π₯ = 1+2π₯ π₯β1
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Test Your Understanding
If π π₯ = 2π₯+1 3 , find π β1 (4). If π π₯ = π₯ 2π₯β1 , find π β1 (π₯). ? ? π= ππ+π π ππ=ππ+π ππβπ=ππ π= ππβπ π π βπ π = ππβπ π π βπ π = ππ π π= π ππβπ πππβπ=π πππβπ=π π ππβπ =π π= π ππβπ π βπ π = π ππβπ
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Exercise 2 ? ? ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet)
Find π β1 (π₯) for the following functions. π π₯ =5π₯ π βπ π = π π π π₯ =1+π₯ π βπ π =πβπ π π₯ =6π₯β π βπ π = π+π π π π₯ = π₯ π βπ π =ππβπ π π₯ =5 π₯ π βπ π = πβπ π π π π₯ =10β3π₯ π βπ π = ππβπ π [Edexcel IGCSE Jan2016(R)-3H Q16c] π π₯ = 2π₯ π₯β1 Find π β1 π₯ = π πβπ Find π β1 (π₯) for the following functions. π π₯ = π₯ π₯ π βπ π = ππ πβπ π π₯ = π₯β2 π₯ π βπ π = π πβπ π π₯ = 2π₯β1 π₯β π βπ π = πβπ πβπ π π₯ = 1βπ₯ 3π₯ π βπ π = πβπ ππ+π π π₯ = 3π₯ 3+2π₯ π βπ π = ππ πβππ Find the value of π for which π π₯ = π₯ π₯+π is a self inverse function. π βπ π = ππ πβπ If self-inverse: π π+π β‘ ππ πβπ π π π + π π πβ‘πβ π π For π π and π terms to match, π=βπ. 1 3 ? ? a a b ? ? b c ? ? d ? c ? e ? d f ? ? e 2 N ? ?
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ππ 2 = π π₯ =3π₯+1 π π₯ = π₯ 2 Composite Functions 49? 13?
Have a guess! (Click your answer) ππ 2 = 49? 13? ππ(2) means π π 2 , i.e. βπ of π of 2β. We therefore apply the functions to the input in sequence from right to left.
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π π₯ =3π₯+1 π π₯ = π₯ 2 Examples ? ? ? ? ππ 5 =π π π =π ππ =ππ
Bro Tip: I highly encourage you to write this first. It will help you when you come to the algebraic ones. Determine: ππ 5 =π π π =π ππ =ππ ππ β1 =π π βπ =π βπ =π ππ 4 =π π π =π ππ =ππ ππ π₯ =π π π =π ππ+π = ππ+π π ? ? ? Bro Note: This can also be written as π 2 (π₯), but you wonβt encounter this notation in GCSE/IGCSE FM. ?
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π π₯ =2π₯+1 π π₯ = 1 π₯ More Algebraic Examples ? ? ? ? ?
Determine: ππ π₯ =π π π =π π π =π π π +π= π π +π ππ π₯ =π ππ+π = π ππ+π ππ π₯ =π ππ+π =π ππ+π +π=ππ+π ππ π₯ =π π π = π π π =π ? ? ? ? ?
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Test Your Understanding
If π π₯ = 2 π₯+1 and π π₯ = π₯ 2 β1, determine ππ(π₯). π π π βπ = π π π βπ+π = π π π 1 2 ? 3 A function π is such that π π₯ =3π₯+1 The function π is such that π π₯ =π π₯ 2 where π is a constant. Given that ππ 3 =55, determine the value of π. ππ π =π ππ =π ππ +π =πππ+π=ππ π=π ? ππ βπ =π π βπ =π βπ =π ?
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Exercise 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet) 1
If π π₯ =3π₯ and π π₯ =π₯+1, determine: ππ 2 =π ππ 4 =ππ ππ π₯ =ππ+π ππ π₯ =ππ+π ππ π₯ =π+π If π π₯ =2π₯+1 and π π₯ =3π₯+1 determine: ππ π₯ =ππ+π ππ π₯ =ππ+π ππ π₯ =ππ+π If π π₯ = π₯ 2 β2π₯ and π π₯ =π₯+1, find ππ(π₯), simplifying your expression. π π+π = π+π π βπ π+π = π π +ππ+πβππβπ = π π βπ If π π₯ =π₯+π and π π₯ = π₯ 2 and ππ 3 =16, find the possible values of π. ππ π =π π+π = π+π π =ππ π=π,βπ If π π₯ =2(π₯+π) and π π₯ = π₯ 2 βπ₯ and ππ 3 =30, find π. π π π =π π =π π+π =ππ π=π 6 Let π π₯ =π₯+1 and π π₯ = π₯ 2 +1. If ππ π₯ =17, determine the possible values of π₯. ππ π = π+π π +π=ππ π π +ππ+π=ππ π π +ππβππ=π π+π πβπ =π π=βπ ππ π=π Let π π₯ = π₯ 2 +3π₯ and π π₯ =π₯β2. If ππ π₯ =0, determine the possible values of π₯. π=βπ ππ π=π [Based on MAT question] π π₯ =π₯+1 and π π₯ =2π₯ Let π π (π₯) means that you apply the function π π times. a) Find π π (π₯) in terms of π₯ and π. =π+π b) Note that π π 2 π π₯ =4π₯+4. Find all other ways of combining π and π that result in the function 4π₯+4. π π π, π π πππ, π π π π ? ? ? ? ? ? 2 ? ? ? 7 3 ? ? N 4 ? 5 ? ?
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This be ye end of GCSE functions content.
Beyond this point there be IGCSE Further Maths. Yarr.
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#4 :: Piecewise Functions
Sometimes functions are defined in βpiecesβ, with a different function for different ranges of π₯ values. Sketch > Sketch > Sketch > (2, 9) (0, 5) (-1, 0) (5, 0)
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Test Your Understanding
π π₯ = π₯ 2 0β€π₯<1 1 1β€π₯<2 3βπ₯ 2β€π₯<3 Sketch Sketch Sketch This example was used on the specification itself! (2, 1) (1, 1) (3, 0)
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Exercise 4 b ? c ? ? a ? (Exercises on provided sheet)
[Jan 2013 Paper 2] A function π(π₯) is defined as: π π₯ = 4 π₯<β2 π₯ 2 β2β€π₯β€2 12β4π₯ π₯>2 Draw the graph of π¦=π(π₯) for β4β€π₯β€4 Use your graph to write down how many solutions there are to π π₯ = sols Solve π π₯ =β ππβππ=βππ βπ= ππ π [June 2013 Paper 2] A function π(π₯) is defined as: π π₯ = π₯+3 β3β€π₯<0 3 0β€π₯<1 5β2π₯ 1β€π₯β€2 Draw the graph of π¦=π(π₯) for β3β€π₯<2 1 2 b ? ? c ? a ?
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Exercise 4 ? Sketch ? ? (Exercises on provided sheet)
[Specimen 1 Q4] A function π(π₯) is defined as: π π₯ = 3π₯ 0β€π₯<1 3 1β€π₯<3 12β3π₯ 3β€π₯β€4 Calculate the area enclosed by the graph of π¦=π π₯ and the π₯βaxis. 3 [Set 1 Paper 1] A function π(π₯) is defined as: π π₯ = 3 0β€π₯<2 π₯+1 2β€π₯<4 9βπ₯ 4β€π₯β€9 Draw the graph of π¦=π(π₯) for 0β€π₯β€9. 4 ? Sketch ? Area = π ?
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Exercise 4 ? ? (Exercises on provided sheet) [AQA Worksheet Q9] 6
π π₯ = β π₯ 2 0β€π₯<2 β4 2β€π₯<3 2π₯β10 3β€π₯β€5 Draw the graph of π(π₯) from 0β€π₯β€5. 6 [AQA Worksheet Q10] π π₯ = 2π₯ 0β€π₯<1 3βπ₯ 1β€π₯<4 π₯β7 3 4β€π₯β€7 5 ? 2 -1 -2 -3 -4 3 7 -1 Show that ππππ ππ π΄:ππππ ππ π΅= 3:2 Area of π¨= π π ΓπΓπ=π Area of π©= π π ΓπΓπ=π ?
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Domain and Range 1 -1 π π₯ = π₯ 2 2.89 1.7 4 2 3.1 9.61 ... ... Inputs
Outputs π 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.
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Example π π₯ = π₯ 2 Sketch: for all π₯ Suitable Domain: Range: π π₯ β₯0 ? ?
π¦ π π₯ = π₯ 2 Sketch: π₯ Bro Note: By βsuitableβ, 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 βproperβ maths weβd use π₯ββ to mean βπ₯ can be any element in the set of real numbersβ, but the syllabus is looking for βfor all π₯β. Range: ? π π₯ β₯0 Look at the π¦ values on the graph. The output has to be positive, since itβs been squared. B Bro Tip: Note that the domain is in terms of π₯ and the range in terms of π π₯ .
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Test Your Understanding
? π¦ Sketch: π π₯ = π₯ π₯ Suitable Domain: π₯β₯0 ? Presuming the output has to be a real number, we canβt input negative numbers into our function. Range: ? π π₯ β₯0 The output, again, can only be positive.
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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 π π₯ = 1 π₯ Domain For all π₯ except 0 Range For all π π₯ except 0 Function π π₯ =2π₯ Domain For all π₯ Range For all π(π₯) Function π π₯ = 2 π₯ Domain For all π₯ Range π π₯ >0 ? ? ? 4 5 6 Function π π₯ = sin π₯ Domain For all π₯ Range β1β€π π₯ β€1 Function π π₯ =2 cos π₯ Domain For all π₯ Range β2β€π π₯ β€2 Function π π₯ = π₯ 3 +1 Domain For all π₯ Range For all π(π₯) ? ? ? 7 8 Function π π₯ = 1 π₯β2 +1 Domain For all π₯ except 2 Range For all π π₯ except 1 Function π π₯ = 2cos π₯+1 Domain π₯>β1 Range β2β€π π₯ β€2 ? ?
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Range of Quadratics A common exam question is to determine the range of a quadratic. The sketch shows the function π¦=π(π₯) where π π₯ = π₯ 2 β4π₯+7. Determine the range of π(π₯). π¦ ? We need the minimum point, since from the graph we can see that π (i.e. π(π)) can be anything greater than this. π π = πβπ π +π The minimum point is (π,π) thus the range is: π π β₯π (note the β₯ rather than >) 3 π₯ An alternative way of thinking about it, once youβve completed the square, is that anything squared is at least 0. So if π₯β2 3 is at least 0, then clearly π₯β is at least 3.
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Test Your Understanding
π¦ The sketch shows the function π¦=π(π₯) where π π₯ = π₯+2 π₯β4 . Determine the range of π(π₯). ? π₯+2 π₯β4 = π₯ 2 β2π₯β = π₯β1 2 β9 Therefore π π₯ β₯β9 π₯ 1,β9 π¦ The sketch shows the function π¦=π(π₯) where π π₯ =21+4π₯β π₯ 2 . Determine the range of π(π₯). 2,25 ? β π₯ 2 β4π₯β21 =β π₯β2 2 β4β21 =25β π₯β2 2 Therefore π π₯ β€25 π₯
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Range for Restricted Domains
Some questions are a bit jammy by restricting the domain. Look out for this, because it affects the domain! π π₯ = π₯ 2 +4π₯+3, π₯β₯1 Determine the range of π(π₯). ? π¦ Notice how the domain is πβ₯π. π π =(π+π)(π+π) When π=π, π= π π +π+π=π Sketching the graph, we see that when π=π, the function is increasing. Therefore when πβ₯π, π π β₯π π₯ β3 β1 1
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Test Your Understanding
π π₯ = π₯ 2 β3, π₯β€β2 Determine the range of π(π₯). π π₯ =3π₯β2, 0β€π₯<4 Determine the range of π(π₯). ? ? π¦ When π₯=0, π π₯ =β2 When π₯=4, π π₯ =10 Range: βπβ€π π <ππ π₯ β2 When π₯=β2, π π₯ =1 As π₯ decreases from -2, π(π₯) is increasing. Therefore: π π₯ β₯1
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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) β1β€π π₯ β€1 180β€π₯β€360 β1β€π π₯ β€0 0β€π₯β€180 0β€π π₯ β€1 ? ? ?
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Range of Piecewise Functions
Itβs a simple case of just sketching the full function. The sketch shows the graph of π¦=π(π₯) with the domain 0β€π₯β€9 π π₯ = 3 0β€π₯<2 π₯+1 2β€π₯<4 9βπ₯ 4β€π₯β€9 Determine the range of π(π₯). Graph ? Range ? Range: πβ€π π β€π
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Test Your Understanding
The function π(π₯) is defined for all π₯: π π₯ = 4 π₯<β2 π₯ 2 β2β€π₯β€2 12β4π₯ π₯>2 Determine the range of π(π₯). Graph ? Range ? Range: π π β€π
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Exercise 5 ? ? ? ? ? ? ? ? ? ? ? (exercises on provided sheet)
Work out the range for each of these functions. (a) π π₯ = π₯ for all π₯ π π β₯π (b) π π₯ =3π₯β5, β2β€π₯β€6 βππβ€π π β€ππ (c) π π₯ =3 π₯ 4 , π₯<β2 π π >ππ (a) π π₯ = π₯+2 π₯β3 Give a reason why π₯>0 is not a suitable domain for π(π₯). It would include 3, for which π(π) is undefined. (b) Give a possible domain for π π₯ = π₯β πβ₯π π π₯ =3β2π₯, π<π₯<π The range of π(π₯) is β5<π π₯ <5 Work out π and π. π=βπ, π=π 4 [Set 1 Paper 2] (a) The function π(π₯) is defined as: π π₯ =22β7π₯, β2β€π₯β€π The range of π(π₯) is β13β€π π₯ β€36 Work out the value of π. π=π (b) The function π(π₯) is defined as π π₯ = π₯ 2 β4π₯+5 for all π₯. (i) Express π(π₯) in the form π₯βπ 2 +π π π = πβπ π +π (ii) Hence write down the range of π(π₯). π π β₯π [June 2012 Paper 1] π π₯ =2 π₯ 2 +7 for all values of π₯. (a) What is the value of π β1 ? π βπ =π (b) What is the range of π(π₯)? π π β₯π 1 ? ? ? ? 2 ? ? ? 5 ? ? 3 ? ?
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Exercise 5 ? ? ? ? ? (exercises on provided sheet)
[Jan 2013 Paper 2] π π₯ = sin π₯ Β°β€π₯β€360Β° π π₯ = cos π₯ Β°β€π₯β€π (a) What is the range of π(π₯)? βπβ€π π β€π (b) You are given that 0β€π π₯ β€1. Work out the value of π. π½=ππΒ° By completing the square or otherwise, determine the range of the following functions: (a) π π₯ = π₯ 2 β2π₯+5, for all π₯ = πβπ π +π Range: π π β₯π (b) π π₯ = π₯ 2 +6π₯β2, for all π₯ = π+π π βππ Range: π π β₯βππ 6 8 ? ? Here is a sketch of π π₯ = π₯ 2 +6π₯+π for all π₯, where π is a constant. The range of π(π₯) is π π₯ β₯11. Work out the value of π. π π = π+π π βπ+π βπ+π=ππ π=ππ 7 ? ? ?
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Exercise 5 ? ? ? ? ? (exercises on provided sheet) 9 10
The straight line shows a sketch of π¦=π(π₯) for the full domain of the function. (a) State the domain of the function. πβ€π π β€ππ (b) Work out the equation of the line. π π =βππ+ππ π(π₯) is a quadratic function with domain all real values of π₯. Part of the graph of π¦=π π₯ is shown. (a) Write down the range of π(π₯). π π β€π (b) Use the graph to find solutions of the equation π π₯ =1. π=βπ.π, π.π (c) Use the graph to solve π π₯ <0. π<βπ ππ π>π ? ? ? ? ?
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Exercise 5 ? ? ? (exercises on provided sheet)
The function π(π₯) is defined as: π π₯ = π₯ 2 β4 0β€π₯<3 14β3π₯ 3β€π₯β€5 Work out the range of π π₯ . π(π)β€π The function π(π₯) has the domain β3β€π₯β€3 and is defined as: π π₯ = π₯ 2 +3π₯+2 β3β€π₯<0 2+π₯ 0β€π₯β€3 Work out the range of π π₯ . β π π β€π π β€π 11 13 ? 12 [June 2012 Paper 2] A sketch of π¦=π(π₯) for domain 0β€π₯β€8 is shown. The graph is symmetrical about π₯=4. The range of π(π₯) is 0β€π π₯ β€12. Work out the function π(π₯). π π₯ = ? 0β€π₯β€4 ? 4<π₯β€8 π π = ππ πβ€πβ€π ππβππ π<πβ€π ? ?
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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= π π =π πβπ=π πβπ π=ππ+π π π =ππ+π π¦ ? 11 ? 3 π₯ 1 5
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Constructing a function from a domain/range
Sometimes thereβs the additional constraint that the function is βincreasingβ or βdecreasingβ. Weβll cover this in more depth when we do calculus, but the meaning of these words should be obvious. π π₯ is a decreasing function with domain 4β€π₯β€6 and range 7β€π π₯ β€19. ? π¦ π=β ππ π =βπ πβπ=βπ πβπ π=βππ+ππ π π =ππβππ 19 7 π₯ 4 6
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Exercise 6 (exercises on provided sheet) 1 Domain is 1β€π₯<3. Range 1β€π π₯ β€3. π(π₯) is an increasing function. π π =π Domain is 1β€π₯β€3. Range 1β€π π₯ β€3. π(π₯) is a decreasing function. π π =ππβπ Domain is 5β€π₯β€7. Range 7β€π π₯ β€11. π(π₯) is an increasing function. π π =πππβπ Domain is 5β€π₯β€7. Range 7β€π π₯ β€11. π(π₯) is a decreasing function. π π =ππβππ Domain is β4β€π₯β€7. Range 4β€π π₯ β€8. π(π₯) is a decreasing function. π π = ππ ππ β π ππ π ? 2 ? 3 ? 4 ? 5 ?
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