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Functions Inverses
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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 π¦= 1 π₯β2 π¦= 1 π₯β2 π₯β4 π¦= π₯β4
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An Inverse function is an βoppositeβ function
Notes An Inverse function is an βoppositeβ function e.g. The inverse of f (x) is written as: f-1(x) and
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π β1 π₯ =2(π₯+8) π β1 π₯ = π₯β7 π β1 0 =16 π β1 7 =0 π β1 4 =24
WB1 f and g are functions defined by π π₯ = 1 2 π₯β8 and π π₯ = π₯ 2 +7 Find the inverse functions π β1 (π₯) and π β1 (π₯) and find a) π β1 (0) π β1 (4) π β1 (2π₯) and b) π β1 (7) π β1 (15) π β1 π₯ 3 Γ 1 2 β8 +8 Γ·2 π₯ 2 +7 β7 π₯ π β1 π₯ =2(π₯+8) π β1 π₯ = π₯β7 π β1 0 =16 π β1 7 =0 π β1 4 =24 π β1 15 =2 2 π β1 π₯ 3 = π₯ 3 β7 = π₯β21 3 π β1 2π₯ =2(2π₯+8)=4(π₯+4)
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π₯= π¦ 2 3 β5 β π¦ 2 =3(π₯+5) β π¦= 3(π₯+5) π β1 π₯ = 3(π₯+5) , π₯β₯β5
WB2a SWAPPING METHOD Find the inverse of each of these functions π π₯ = π₯ 2 3 β5 xβπ
b) π π₯ = 6 3π₯β c) β π₯ = 3π₯+2 2π₯β5 π₯= π¦ 2 3 β5 Step 1 swap x and y around Step 2 rearrange until you get y = β¦ β π¦ 2 =3(π₯+5) β π¦= 3(π₯+5) Step 3 write in function notation, write the domain if you know it π β1 π₯ = 3(π₯+5) , π₯β₯β5
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π₯ 3π¦β2 =6 β 3π₯π¦β2π₯=6 β π¦= 2π₯+6 3π₯ β 3π¦π₯=2x+6 π β1 π₯ = 2π₯+6 3π₯ , π₯β 0
WB2b Find the inverse of each of these functions a) π π₯ = π₯ 2 3 β5 xβπ
b) π π₯ = 6 3π₯β c) β π₯ = 3π₯+2 2π₯β5 π₯= 6 3π¦β2 Step 1 swap x and y around Step 2 rearrange until you get y = β¦ π₯ 3π¦β2 =6 β 3π₯π¦β2π₯=6 β 3π¦π₯=2x+6 β π¦= 2π₯+6 3π₯ Step 3 write in function notation, write the domain if you know it π β1 π₯ = 2π₯+6 3π₯ , π₯β 0
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β 2π₯π¦β3π¦=5π₯+2 β π₯ 2π¦β5 =3π¦+2 β π¦= 5π₯+2 2π₯β3 β π¦ 2π₯β3 =5π₯+2
WB2c Find the inverse of each of these functions a) π π₯ = π₯ 2 3 β5 xβπ
b) π π₯ = 6 3π₯β c) β π₯ = 3π₯+2 2π₯β5 π₯= 3π¦+2 2π¦β5 Step 1 swap x and y around Step 2 rearrange until you get y = β¦ β π₯ 2π¦β5 =3π¦+2 β 2π₯π¦β3π¦=5π₯+2 β π¦= 5π₯+2 2π₯β3 β π¦ 2π₯β3 =5π₯+2 Step 3 write in function notation, write the domain if you know it β β1 π₯ = 5π₯+2 2π₯β3 , π₯β 3 2
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β 2π¦+1=ππππ ππ 3(π₯+5) β3(π₯+5)= sin (2π¦+1) β π¦= ππππ ππ 3π₯+15 β1 2
WB 3a Find the inverse of each of these functions a) π π₯ = sin (2π₯+1) 3 β5 xβπ
b) π π₯ = π 3π₯β c) β π₯ = ln 4β6π₯ π₯= sin (2π¦+1) 3 β5 Step 1 swap x and y around Step 2 rearrange until you get y = β¦ β3(π₯+5)= sin (2π¦+1) β 2π¦+1=ππππ ππ 3(π₯+5) β π¦= ππππ ππ 3π₯+15 β1 2 Step 3 write in function notation, write the domain if you know it π β1 π₯ = ππππ ππ 3π₯+15 β1 2 , β1β€(3π₯+15)β€1 β14 3 β€π₯β€β 16 3
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π₯= π 3π¦β2 β ln π₯ =3π¦β2 β π¦= 1 3 2+ ln π₯ π β1 π₯ = 1 3 2+ ln π₯ , π₯>0
WB 3b Find the inverse of each of these functions a) π π₯ = sin (2π₯+1) 3 β5 xβπ
b) π π₯ = π 3π₯β c) β π₯ = ln 4β6π₯ π₯= π 3π¦β2 Step 1 swap x and y around Step 2 rearrange until you get y = β¦ β ln π₯ =3π¦β2 β π¦= ln π₯ Step 3 write in function notation, write the domain if you know it π β1 π₯ = ln π₯ , π₯>0
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π₯= ln 4β6π¦ β π π₯ =4 β6π¦ β π¦= 1 6 4β π π₯ π β1 π₯ = 1 6 4β π π₯ , π₯βπ
WB 3c Find the inverse of each of these functions a) π π₯ = sin (2π₯+1) 3 β5 xβπ
b) π π₯ = π 3π₯β c) β π₯ = ln 4β6π₯ π₯= ln 4β6π¦ Step 1 swap x and y around Step 2 rearrange until you get y = β¦ β π π₯ =4 β6π¦ β π¦= β π π₯ Step 3 write in function notation, write the domain if you know it π β1 π₯ = β π π₯ , π₯βπ
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π) π π₯ = (π₯+3) 2 β1 b) π π₯ =3 (π₯β2) 2 +6 Let π₯= (π¦+3) 2 β1
WB 4 Rearrange each function to complete square form then find its inverse a) π π₯ = π₯ 2 +6π₯+8 xβπ
b) π π₯ = 3π₯ 2 β12π₯+18 π) π π₯ = (π₯+3) 2 β1 b) π π₯ =3 (π₯β2) 2 +6 Let π₯= (π¦+3) 2 β1 Let π₯=3 (π¦β2) 2 +6 (π¦+3) 2 =π₯+1 (π¦β2) 2 = π₯β6 3 π¦+3= π₯+1 π¦β2= π₯β6 3 π¦=β3+ π₯+1 π β1 π₯ =2+ π₯β6 3 , π₯β₯6 π β1 π₯ =β3+ π₯+1 , π₯β₯β1
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Think Pair π π₯ =2π₯+1 and π β1 π₯ = π₯β1 2 share Use graph software
WB 5 Sketch each pair of functions on the same set of axes π π₯ =2π₯+1 and π β1 π₯ = π₯β1 2 π π₯ = π₯ and π β1 π₯ = π₯ Use graph software What do you notice?
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π π₯ Domain π<π Range π π πΉ
Notes π π₯ Domain π<π Range π π πΉ Domain of f is range of its inverse β¦ range of f is domain of its inverse π β1 π₯ Domain π π πΉ Range π<π
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a) β ln π¦ π₯ = 4π‘ b) β π 2π₯ = π¦ 5 β 2π₯= ln π¦ 5 β y x = π 4π‘
Think Pair Share WB a) Make x the subject of: πππ¦ βπππ₯=4π‘ b) Make x the subject of: π¦ = 5π 2π₯ a) β ln π¦ π₯ = 4π‘ b) β π 2π₯ = π¦ 5 β π₯= ln π¦ 5 β y x = π 4π‘ β π₯= ln π¦ 5 β π¦= π₯ π 4π‘ β π₯=π¦ π β4π‘
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WB7 The function f is defined by π:π₯β ln 3π₯β2 , π₯ββ, π₯β₯ 2 3
Find π β1 (π₯) State the domain of π β1 (π₯) π¦= 2 3 π π₯ = ln 3π₯β2 Let π₯= ln 3π¦β2 π π₯ = 3π¦ β2 π¦ = π π₯ +2 3 π β1 (π₯) = π π₯ +2 3 Domain π₯ββ
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WB8 The function f is defined by π:π₯β 2π π₯+1
Find π β1 (π₯) State the domain of π β1 (π₯) π π₯ = 2π π₯+1 Let π₯= 2π π¦+1 π π¦+1 = π₯ 2 Domain π₯ββ, π₯>0 π¦+1= ln π₯ 2 π β1 π₯ = ln π₯ 2 β1 Domain π₯ββ, π₯>0
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a) = 4 π₯+7 π₯+7 + 3 π₯+7 = 4π₯+31 π₯+7 b) Let x= 4π¦+31 π¦+7 β x y+7 =4π¦+31
WB9 exam Q π π₯ =4+ 3 π₯ xβπ
, π₯β 7 a) Express 4+ 3 π₯+7 s a single fraction Find an expression for π β1 π₯ write the domain of π β1 π₯ a) = 4 π₯+7 π₯ π₯+7 = 4π₯+31 π₯+7 b) Let x= 4π¦+31 π¦+7 β x y+7 =4π¦+31 β π₯π¦β4π¦=31β7π₯ β π¦= 31β7π₯ π₯β4 c) Domain π₯βπ
π₯β 4
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a) Let x=1+ 3 π¦β3 β (π₯β1) 2 = 3 π¦β3 β π¦β3= 3 (π₯β1) 2
WB10 exam Q π π₯ = π₯β xβπ
, π₯β 3 a) Find an expression for π β1 π₯ write the domain of π β1 π₯ a) Let x= π¦β3 β (π₯β1) 2 = 3 π¦β3 β π¦β3= 3 (π₯β1) 2 β π β1 π₯ = 3 (π₯β1) 2 +3 b) Domain π₯βπ
π₯β 1
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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 β One thing to improve is β
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