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Functions Inverses.

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1 Functions Inverses

2 Starter: sketch each function
Functions Inverses KUS objectives BAT Find the inverse of functions BAT understand the inverse of a function graphically Starter: sketch each function 𝑦= 1 π‘₯βˆ’2 𝑦= 1 π‘₯βˆ’2 π‘₯βˆ’4 𝑦= π‘₯βˆ’4

3 An Inverse function is an β€˜opposite’ function
Notes An Inverse function is an β€˜opposite’ function e.g. The inverse of f (x) is written as: f-1(x) and

4 𝑓 βˆ’1 π‘₯ =2(π‘₯+8) 𝑔 βˆ’1 π‘₯ = π‘₯βˆ’7 𝑓 βˆ’1 0 =16 𝑔 βˆ’1 7 =0 𝑓 βˆ’1 4 =24
WB1 f and g are functions defined by 𝑓 π‘₯ = 1 2 π‘₯βˆ’8 and 𝑔 π‘₯ = π‘₯ 2 +7 Find the inverse functions 𝑓 βˆ’1 (π‘₯) and 𝑔 βˆ’1 (π‘₯) and find a) 𝑓 βˆ’1 (0) 𝑓 βˆ’1 (4) 𝑓 βˆ’1 (2π‘₯) and b) 𝑔 βˆ’1 (7) 𝑔 βˆ’1 (15) 𝑔 βˆ’1 π‘₯ 3 Γ— 1 2 βˆ’8 +8 Γ·2 π‘₯ 2 +7 βˆ’7 π‘₯ 𝑓 βˆ’1 π‘₯ =2(π‘₯+8) 𝑔 βˆ’1 π‘₯ = π‘₯βˆ’7 𝑓 βˆ’1 0 =16 𝑔 βˆ’1 7 =0 𝑓 βˆ’1 4 =24 𝑔 βˆ’1 15 =2 2 𝑔 βˆ’1 π‘₯ 3 = π‘₯ 3 βˆ’7 = π‘₯βˆ’21 3 𝑓 βˆ’1 2π‘₯ =2(2π‘₯+8)=4(π‘₯+4)

5 π‘₯= 𝑦 2 3 βˆ’5 β‡’ 𝑦 2 =3(π‘₯+5) β‡’ 𝑦= 3(π‘₯+5) 𝑓 βˆ’1 π‘₯ = 3(π‘₯+5) , π‘₯β‰₯βˆ’5
WB2a SWAPPING METHOD Find the inverse of each of these functions 𝑓 π‘₯ = π‘₯ 2 3 βˆ’5 xβˆˆπ‘… b) 𝑔 π‘₯ = 6 3π‘₯βˆ’ c) β„Ž π‘₯ = 3π‘₯+2 2π‘₯βˆ’5 π‘₯= 𝑦 2 3 βˆ’5 Step 1 swap x and y around Step 2 rearrange until you get y = … β‡’ 𝑦 2 =3(π‘₯+5) β‡’ 𝑦= 3(π‘₯+5) Step 3 write in function notation, write the domain if you know it 𝑓 βˆ’1 π‘₯ = 3(π‘₯+5) , π‘₯β‰₯βˆ’5

6 π‘₯ 3π‘¦βˆ’2 =6 β‡’ 3π‘₯π‘¦βˆ’2π‘₯=6 β‡’ 𝑦= 2π‘₯+6 3π‘₯ β‡’ 3𝑦π‘₯=2x+6 𝑔 βˆ’1 π‘₯ = 2π‘₯+6 3π‘₯ , π‘₯β‰ 0
WB2b Find the inverse of each of these functions a) 𝑓 π‘₯ = π‘₯ 2 3 βˆ’5 xβˆˆπ‘… b) 𝑔 π‘₯ = 6 3π‘₯βˆ’ c) β„Ž π‘₯ = 3π‘₯+2 2π‘₯βˆ’5 π‘₯= 6 3π‘¦βˆ’2 Step 1 swap x and y around Step 2 rearrange until you get y = … π‘₯ 3π‘¦βˆ’2 =6 β‡’ 3π‘₯π‘¦βˆ’2π‘₯=6 β‡’ 3𝑦π‘₯=2x+6 β‡’ 𝑦= 2π‘₯+6 3π‘₯ Step 3 write in function notation, write the domain if you know it 𝑔 βˆ’1 π‘₯ = 2π‘₯+6 3π‘₯ , π‘₯β‰ 0

7 β‡’ 2π‘₯π‘¦βˆ’3𝑦=5π‘₯+2 β‡’ π‘₯ 2π‘¦βˆ’5 =3𝑦+2 β‡’ 𝑦= 5π‘₯+2 2π‘₯βˆ’3 β‡’ 𝑦 2π‘₯βˆ’3 =5π‘₯+2
WB2c Find the inverse of each of these functions a) 𝑓 π‘₯ = π‘₯ 2 3 βˆ’5 xβˆˆπ‘… b) 𝑔 π‘₯ = 6 3π‘₯βˆ’ c) β„Ž π‘₯ = 3π‘₯+2 2π‘₯βˆ’5 π‘₯= 3𝑦+2 2π‘¦βˆ’5 Step 1 swap x and y around Step 2 rearrange until you get y = … β‡’ π‘₯ 2π‘¦βˆ’5 =3𝑦+2 β‡’ 2π‘₯π‘¦βˆ’3𝑦=5π‘₯+2 β‡’ 𝑦= 5π‘₯+2 2π‘₯βˆ’3 β‡’ 𝑦 2π‘₯βˆ’3 =5π‘₯+2 Step 3 write in function notation, write the domain if you know it β„Ž βˆ’1 π‘₯ = 5π‘₯+2 2π‘₯βˆ’3 , π‘₯β‰  3 2

8 β‡’ 2𝑦+1=π‘Žπ‘Ÿπ‘π‘ π‘–π‘› 3(π‘₯+5) β‡’3(π‘₯+5)= sin (2𝑦+1) β‡’ 𝑦= π‘Žπ‘Ÿπ‘π‘ π‘–π‘› 3π‘₯+15 βˆ’1 2
WB 3a Find the inverse of each of these functions a) 𝑓 π‘₯ = sin (2π‘₯+1) 3 βˆ’5 xβˆˆπ‘… b) 𝑔 π‘₯ = 𝑒 3π‘₯βˆ’ c) β„Ž π‘₯ = ln 4βˆ’6π‘₯ π‘₯= sin (2𝑦+1) 3 βˆ’5 Step 1 swap x and y around Step 2 rearrange until you get y = … β‡’3(π‘₯+5)= sin (2𝑦+1) β‡’ 2𝑦+1=π‘Žπ‘Ÿπ‘π‘ π‘–π‘› 3(π‘₯+5) β‡’ 𝑦= π‘Žπ‘Ÿπ‘π‘ π‘–π‘› 3π‘₯+15 βˆ’1 2 Step 3 write in function notation, write the domain if you know it 𝑓 βˆ’1 π‘₯ = π‘Žπ‘Ÿπ‘π‘ π‘–π‘› 3π‘₯+15 βˆ’1 2 , βˆ’1≀(3π‘₯+15)≀1 βˆ’14 3 ≀π‘₯β‰€βˆ’ 16 3

9 π‘₯= 𝑒 3π‘¦βˆ’2 β‡’ ln π‘₯ =3π‘¦βˆ’2 β‡’ 𝑦= 1 3 2+ ln π‘₯ 𝑔 βˆ’1 π‘₯ = 1 3 2+ ln π‘₯ , π‘₯>0
WB 3b Find the inverse of each of these functions a) 𝑓 π‘₯ = sin (2π‘₯+1) 3 βˆ’5 xβˆˆπ‘… b) 𝑔 π‘₯ = 𝑒 3π‘₯βˆ’ c) β„Ž π‘₯ = ln 4βˆ’6π‘₯ π‘₯= 𝑒 3π‘¦βˆ’2 Step 1 swap x and y around Step 2 rearrange until you get y = … β‡’ ln π‘₯ =3π‘¦βˆ’2 β‡’ 𝑦= ln π‘₯ Step 3 write in function notation, write the domain if you know it 𝑔 βˆ’1 π‘₯ = ln π‘₯ , π‘₯>0

10 π‘₯= ln 4βˆ’6𝑦 β‡’ 𝑒 π‘₯ =4 βˆ’6𝑦 β‡’ 𝑦= 1 6 4βˆ’ 𝑒 π‘₯ 𝑔 βˆ’1 π‘₯ = 1 6 4βˆ’ 𝑒 π‘₯ , π‘₯βˆˆπ‘…
WB 3c Find the inverse of each of these functions a) 𝑓 π‘₯ = sin (2π‘₯+1) 3 βˆ’5 xβˆˆπ‘… b) 𝑔 π‘₯ = 𝑒 3π‘₯βˆ’ c) β„Ž π‘₯ = ln 4βˆ’6π‘₯ π‘₯= ln 4βˆ’6𝑦 Step 1 swap x and y around Step 2 rearrange until you get y = … β‡’ 𝑒 π‘₯ =4 βˆ’6𝑦 β‡’ 𝑦= βˆ’ 𝑒 π‘₯ Step 3 write in function notation, write the domain if you know it 𝑔 βˆ’1 π‘₯ = βˆ’ 𝑒 π‘₯ , π‘₯βˆˆπ‘…

11 π‘Ž) 𝑓 π‘₯ = (π‘₯+3) 2 βˆ’1 b) 𝑔 π‘₯ =3 (π‘₯βˆ’2) 2 +6 Let π‘₯= (𝑦+3) 2 βˆ’1
WB 4 Rearrange each function to complete square form then find its inverse a) 𝑓 π‘₯ = π‘₯ 2 +6π‘₯+8 xβˆˆπ‘… b) 𝑔 π‘₯ = 3π‘₯ 2 βˆ’12π‘₯+18 π‘Ž) 𝑓 π‘₯ = (π‘₯+3) 2 βˆ’1 b) 𝑔 π‘₯ =3 (π‘₯βˆ’2) 2 +6 Let π‘₯= (𝑦+3) 2 βˆ’1 Let π‘₯=3 (π‘¦βˆ’2) 2 +6 (𝑦+3) 2 =π‘₯+1 (π‘¦βˆ’2) 2 = π‘₯βˆ’6 3 𝑦+3= π‘₯+1 π‘¦βˆ’2= π‘₯βˆ’6 3 𝑦=βˆ’3+ π‘₯+1 𝑔 βˆ’1 π‘₯ =2+ π‘₯βˆ’6 3 , π‘₯β‰₯6 𝑓 βˆ’1 π‘₯ =βˆ’3+ π‘₯+1 , π‘₯β‰₯βˆ’1

12 Think Pair 𝑓 π‘₯ =2π‘₯+1 and 𝑓 βˆ’1 π‘₯ = π‘₯βˆ’1 2 share Use graph software
WB 5 Sketch each pair of functions on the same set of axes 𝑓 π‘₯ =2π‘₯+1 and 𝑓 βˆ’1 π‘₯ = π‘₯βˆ’1 2 𝑔 π‘₯ = π‘₯ and 𝑔 βˆ’1 π‘₯ = π‘₯ Use graph software What do you notice?

13 𝑓 π‘₯ Domain 𝒙<πŸ– Range π’š 𝝐 𝑹
Notes 𝑓 π‘₯ Domain 𝒙<πŸ– Range π’š 𝝐 𝑹 Domain of f is range of its inverse … range of f is domain of its inverse 𝑓 βˆ’1 π‘₯ Domain 𝒙 𝝐 𝑹 Range π’š<πŸ–

14 a) β‡’ ln 𝑦 π‘₯ = 4𝑑 b) β‡’ 𝑒 2π‘₯ = 𝑦 5 β‡’ 2π‘₯= ln 𝑦 5 β‡’ y x = 𝑒 4𝑑
Think Pair Share WB a) Make x the subject of: 𝑙𝑛𝑦 βˆ’π‘™π‘›π‘₯=4𝑑 b) Make x the subject of: 𝑦 = 5𝑒 2π‘₯ a) β‡’ ln 𝑦 π‘₯ = 4𝑑 b) β‡’ 𝑒 2π‘₯ = 𝑦 5 β‡’ π‘₯= ln 𝑦 5 β‡’ y x = 𝑒 4𝑑 β‡’ π‘₯= ln 𝑦 5 β‡’ 𝑦= π‘₯ 𝑒 4𝑑 β‡’ π‘₯=𝑦 𝑒 βˆ’4𝑑

15 WB7 The function f is defined by 𝑓:π‘₯β†’ ln 3π‘₯βˆ’2 , π‘₯βˆˆβ„›, π‘₯β‰₯ 2 3
Find 𝑓 βˆ’1 (π‘₯) State the domain of 𝑓 βˆ’1 (π‘₯) 𝑦= 2 3 𝑓 π‘₯ = ln 3π‘₯βˆ’2 Let π‘₯= ln 3π‘¦βˆ’2 𝑒 π‘₯ = 3𝑦 βˆ’2 𝑦 = 𝑒 π‘₯ +2 3 𝑓 βˆ’1 (π‘₯) = 𝑒 π‘₯ +2 3 Domain π‘₯βˆˆβ„›

16 WB8 The function f is defined by 𝑓:π‘₯β†’ 2𝑒 π‘₯+1
Find 𝑓 βˆ’1 (π‘₯) State the domain of 𝑓 βˆ’1 (π‘₯) 𝑓 π‘₯ = 2𝑒 π‘₯+1 Let π‘₯= 2𝑒 𝑦+1 𝑒 𝑦+1 = π‘₯ 2 Domain π‘₯βˆˆβ„›, π‘₯>0 𝑦+1= ln π‘₯ 2 𝑓 βˆ’1 π‘₯ = ln π‘₯ 2 βˆ’1 Domain π‘₯βˆˆβ„›, π‘₯>0

17 a) = 4 π‘₯+7 π‘₯+7 + 3 π‘₯+7 = 4π‘₯+31 π‘₯+7 b) Let x= 4𝑦+31 𝑦+7 β‡’ x y+7 =4𝑦+31
WB9 exam Q 𝑓 π‘₯ =4+ 3 π‘₯ xβˆˆπ‘…, π‘₯β‰ 7 a) Express 4+ 3 π‘₯+7 s a single fraction Find an expression for 𝑓 βˆ’1 π‘₯ write the domain of 𝑓 βˆ’1 π‘₯ a) = 4 π‘₯+7 π‘₯ π‘₯+7 = 4π‘₯+31 π‘₯+7 b) Let x= 4𝑦+31 𝑦+7 β‡’ x y+7 =4𝑦+31 β‡’ π‘₯π‘¦βˆ’4𝑦=31βˆ’7π‘₯ β‡’ 𝑦= 31βˆ’7π‘₯ π‘₯βˆ’4 c) Domain π‘₯βˆˆπ‘… π‘₯β‰ 4

18 a) Let x=1+ 3 π‘¦βˆ’3 β‡’ (π‘₯βˆ’1) 2 = 3 π‘¦βˆ’3 β‡’ π‘¦βˆ’3= 3 (π‘₯βˆ’1) 2
WB10 exam Q 𝑓 π‘₯ = π‘₯βˆ’ xβˆˆπ‘…, π‘₯β‰ 3 a) Find an expression for 𝑓 βˆ’1 π‘₯ write the domain of 𝑓 βˆ’1 π‘₯ a) Let x= π‘¦βˆ’3 β‡’ (π‘₯βˆ’1) 2 = 3 π‘¦βˆ’3 β‡’ π‘¦βˆ’3= 3 (π‘₯βˆ’1) 2 β‡’ 𝑓 βˆ’1 π‘₯ = 3 (π‘₯βˆ’1) 2 +3 b) Domain π‘₯βˆˆπ‘… π‘₯β‰ 1

19 BAT understand the inverse of a function graphically
KUS objectives BAT Find the inverse of functions BAT understand the inverse of a function graphically self-assess One thing learned is – One thing to improve is –

20 END


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