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Modulus Function.

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Presentation on theme: "Modulus Function."β€” Presentation transcript:

1 Modulus Function

2 State the range of 𝐠𝐟(𝐱)
Functions Modulus KUS objectives BAT Understand and draw graphs of modulus functions BAT Find intersections of graphs, including modulus graphs Starter: 𝒇 𝒙 =𝒙+𝟐, π’™βˆˆβ„› and π’ˆ 𝒙 = πŸ’ π’™βˆ’πŸ‘ , π’™βˆˆβ„›, π’™β‰ πŸ‘ Find 𝐟𝐠(πŸ’) Find 𝐠𝐟(βˆ’πŸ) Find 𝐟𝐠(𝐱) Find 𝐠𝐟(π±βˆ’πŸ) Find 𝐟𝐠(𝟐𝐱) State the range of 𝐠𝐟(𝐱)

3 The modulus function makes values positive. We have two cases:
notes The modulus function makes values positive. We have two cases: i) Modulus of the function Graph: -4 4 x y

4 The modulus function makes values positive. We have two cases:
notes The modulus function makes values positive. We have two cases: ii) Modulus of x within the function f(x) Graph: -4 4 x y What type of function is f(x)?

5 π’‡π’ˆ 𝒙 = 𝒙 +πŸ‘ π’‡π’ˆ 𝒙 = 𝒙+πŸ‘ WB1 given that 𝒇 𝒙 = 𝒙 and π’ˆ 𝒙 =𝒙+πŸ‘
Sketch the graphs of the composite functions π’‡π’ˆ 𝒙 and π’ˆπ’‡ 𝒙 Indicating clearly which is which 3 π’‡π’ˆ 𝒙 = 𝒙 +πŸ‘ -3 3 π’‡π’ˆ 𝒙 = 𝒙+πŸ‘

6 Think Pair Share WB2ab Sketch each pair of graphs on the same axes: 𝑦=3π‘₯βˆ’6 𝑦=3π‘₯+1 𝑦=3 π‘₯ βˆ’6 𝑦= 3π‘₯+1

7 Think Pair Share WB2cd Sketch each pair of graphs on the same axes: 𝑦= π‘₯ 2 βˆ’4 𝑦=4π‘₯βˆ’3 𝑦= π‘₯ 2 βˆ’4 𝑦=4 π‘₯ βˆ’3

8 𝑓(π‘₯)= (π‘₯βˆ’4) 2 βˆ’6 And 𝑓 π‘₯ 𝑦= 1 π‘₯βˆ’3 𝑦= 1 π‘₯ βˆ’3
Think Pair Share WB2ef Sketch each pair of graphs on the same axes: 𝑦= 1 π‘₯βˆ’3 𝑓(π‘₯)= (π‘₯βˆ’4) 2 βˆ’6 𝑦= 1 π‘₯ βˆ’3 And 𝑓 π‘₯

9 two intersections 1st βˆ’ π‘₯βˆ’4 = 1 2 π‘₯+4 π‘₯=0 point 0, 4 2nd π‘₯βˆ’4= 1 2 π‘₯+4
WB 𝑓(π‘₯)= π‘₯βˆ’4 and 𝑔 π‘₯ = 1 2 π‘₯+4 sketch the graphs of each function then find their points of intersection two intersections 1st βˆ’ π‘₯βˆ’4 = 1 2 π‘₯+4 π‘₯=0 point 0, 4 2nd π‘₯βˆ’4= 1 2 π‘₯+4 π‘₯=16 point 16, 12

10 two intersections 1st 2 βˆ’π‘₯ βˆ’3=βˆ’ 1 2 π‘₯+6 π‘₯=βˆ’6 point βˆ’6, 9
WB 𝑓 π‘₯ =2 π‘₯ βˆ’3 and 𝑔 π‘₯ =βˆ’ 1 2 π‘₯+6 sketch the graphs of each function then find their points of intersection two intersections 1st 2 βˆ’π‘₯ βˆ’3=βˆ’ 1 2 π‘₯+6 π‘₯=βˆ’6 point βˆ’6, 9 2nd π‘₯ βˆ’3=βˆ’ 1 2 π‘₯+6 π‘₯= 18 5 = point , 4.2

11 Neither of these gives a correct solution – look at the graph
WB 𝑓(π‘₯)= 1 2 π‘₯βˆ’6 and 𝑔(π‘₯)= 2π‘₯βˆ’4 sketch the graphs of each function then find their points of intersection two intersections 1st βˆ’ 1 2 π‘₯βˆ’6 =2π‘₯βˆ’4 π‘₯=4 point 4, 4 2nd βˆ’ 1 2 π‘₯βˆ’6 =βˆ’(2π‘₯βˆ’4) π‘₯=βˆ’ point βˆ’ 4 3 , 20 3 Note that there are two other possibilities trying to solve algebraically 1 2 π‘₯βˆ’6=2π‘₯βˆ’ and π‘₯βˆ’6=βˆ’(2π‘₯βˆ’4) Neither of these gives a correct solution – look at the graph

12 Put the negative sign on the β€˜easiest’ side π‘₯ 2 βˆ’6π‘₯+8=0
WB6 a) sketch the graph of 𝑦= π‘₯ 2 βˆ’6π‘₯ b) hence, solve the equation π‘₯ 2 βˆ’6π‘₯ =8 b) Four intersections 1st π‘₯ 2 βˆ’6π‘₯=8 π‘₯ 2 βˆ’6π‘₯βˆ’8=0 π‘₯= 6Β± = πŸ‘Β± πŸπŸ• 2nd π‘₯ 2 βˆ’6π‘₯=βˆ’8 Put the negative sign on the β€˜easiest’ side π‘₯ 2 βˆ’6π‘₯+8=0 (π‘₯βˆ’4)(π‘₯βˆ’2)=0 𝒙=𝟐 𝒙=πŸ’

13 WB 𝑓 π‘₯ = 2π‘₯βˆ’6 , π‘₯ πœ– 𝑅 a) Sketch the graph with equation 𝑦=𝑓 π‘₯ showing the coordinates of the points where the graph cuts or meets the axes b) solve 𝑓 π‘₯ =10+π‘₯ 𝑔 π‘₯ = π‘₯ 2 βˆ’2π‘₯+4, π‘₯ πœ– 𝑅, 0≀π‘₯<7 find 𝑓𝑔 5 Find the range of 𝑔 π‘₯ 2π‘₯βˆ’6 =10+π‘₯ 2π‘₯βˆ’6=10+π‘₯ gives π‘₯=16 2π‘₯βˆ’6=βˆ’(10+π‘₯) gives π‘₯=βˆ’ 4 3 𝑐) 𝑓𝑔 5 =𝑓 19 = 2(19)βˆ’6 =32 𝑑) 𝑔 π‘₯ = (π‘₯βˆ’1) 2 + 3, 𝑔 π‘₯ β‰₯3 𝑔 0 =4 , 𝑔 7 =39 range is 3≀ 𝑔 π‘₯ ≀39

14 Hence there is a root in the interval [-4, -3]
WB 𝑓 π‘₯ = π‘₯ 4 βˆ’4π‘₯βˆ’240 Show that there is a root of f(x) = 0 in the interval [-4, -3] Find the coordinates of the turning point on graph of y = f(x) Given that 𝑓 π‘₯ =(π‘₯βˆ’4)( π‘₯ 3 +π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐) find the values of a,b and c Sketch the graph of y = f(x) Hence sketch the graph of y = f(x) 𝑓 βˆ’4 =32 𝑓 βˆ’3 =βˆ’147 Hence there is a root in the interval [-4, -3] 𝑓 β€² (π‘₯)=4 π‘₯ 3 βˆ’4 𝑓 β€² (π‘₯)=0 when π‘₯=1 gives point 1, βˆ’243 c) 𝑓(π‘₯)=(π‘₯βˆ’4)( π‘₯ 3 +4 π‘₯ 2 +16π‘₯+60)

15 g is defined by: f x = π‘₯ 2 βˆ’4π‘₯+11, π‘₯β‰₯0
8 marks 12 mins WB9 Exam Q f is defined by: f x = π‘₯βˆ’4 βˆ’3, π‘₯βˆˆπ‘… g is defined by: f x = π‘₯ 2 βˆ’4π‘₯+11, π‘₯β‰₯0 Solve f x =3 State the range of g(x) find g𝐟 βˆ’πŸ plus a) π’™βˆ’πŸ’ βˆ’πŸ‘=πŸ‘ π‘₯βˆ’4=6, π‘₯=10 π’™βˆ’πŸ’ =6 minus π‘₯βˆ’4=βˆ’6, π‘₯=βˆ’2 b ) 𝒙 𝟐 βˆ’πŸ’π’™+𝟏𝟏= π’™βˆ’πŸ 𝟐 +πŸ• Range is g(x)β‰₯πŸ• 𝐟 βˆ’πŸ = βˆ’πŸ” βˆ’πŸ‘ =πŸ‘ g𝐟 βˆ’πŸ = g πŸ‘ = (πŸ‘) 𝟐 βˆ’πŸ’ πŸ‘ +𝟏𝟏=πŸ–

16 One thing to improve is –
KUS objectives BAT Understand and draw graphs of modulus functions BAT Find intersections of graphs, including modulus graphs self-assess One thing learned is – One thing to improve is –

17 END


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