Modulus Function

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, 𝒙∈ℛ and 𝒈 𝒙 = 𝟒 𝒙−𝟑 , 𝒙∈ℛ, 𝒙≠𝟑 Find 𝐟𝐠(𝟒) Find 𝐠𝐟(−𝟐) Find 𝐟𝐠(𝐱) Find 𝐠𝐟(𝐱−𝟐) Find 𝐟𝐠(𝟐𝐱) State the range of 𝐠𝐟(𝐱)

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    

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)?

𝒇𝒈 𝒙 = 𝒙 +𝟑 𝒇𝒈 𝒙 = 𝒙+𝟑 WB1 given that 𝒇 𝒙 = 𝒙 and 𝒈 𝒙 =𝒙+𝟑 Sketch the graphs of the composite functions 𝒇𝒈 𝒙 and 𝒈𝒇 𝒙 Indicating clearly which is which 3 𝒇𝒈 𝒙 = 𝒙 +𝟑 -3 3 𝒇𝒈 𝒙 = 𝒙+𝟑

Think Pair Share WB2ab Sketch each pair of graphs on the same axes: 𝑦=3𝑥−6 𝑦=3𝑥+1 𝑦=3 𝑥 −6 𝑦= 3𝑥+1

Think Pair Share WB2cd Sketch each pair of graphs on the same axes: 𝑦= 𝑥 2 −4 𝑦=4𝑥−3 𝑦= 𝑥 2 −4 𝑦=4 𝑥 −3

𝑓(𝑥)= (𝑥−4) 2 −6 And 𝑓 𝑥 𝑦= 1 𝑥−3 𝑦= 1 𝑥 −3 Think Pair Share WB2ef Sketch each pair of graphs on the same axes: 𝑦= 1 𝑥−3 𝑓(𝑥)= (𝑥−4) 2 −6 𝑦= 1 𝑥 −3 And 𝑓 𝑥

two intersections 1st − 𝑥−4 = 1 2 𝑥+4 𝑥=0 point 0, 4 2nd 𝑥−4= 1 2 𝑥+4 WB3 𝑓(𝑥)= 𝑥−4 and 𝑔 𝑥 = 1 2 𝑥+4 sketch the graphs of each function then find their points of intersection two intersections 1st − 𝑥−4 = 1 2 𝑥+4 𝑥=0 point 0, 4 2nd 𝑥−4= 1 2 𝑥+4 𝑥=16 point 16, 12

two intersections 1st 2 −𝑥 −3=− 1 2 𝑥+6 𝑥=−6 point −6, 9 WB4 𝑓 𝑥 =2 𝑥 −3 and 𝑔 𝑥 =− 1 2 𝑥+6 sketch the graphs of each function then find their points of intersection two intersections 1st 2 −𝑥 −3=− 1 2 𝑥+6 𝑥=−6 point −6, 9 2nd 2 +𝑥 −3=− 1 2 𝑥+6 𝑥= 18 5 =3.6 point 3.6, 4.2

Neither of these gives a correct solution – look at the graph WB5 𝑓(𝑥)= 1 2 𝑥−6 and 𝑔(𝑥)= 2𝑥−4 sketch the graphs of each function then find their points of intersection two intersections 1st − 1 2 𝑥−6 =2𝑥−4 𝑥=4 point 4, 4 2nd − 1 2 𝑥−6 =−(2𝑥−4) 𝑥=− 4 3 point − 4 3 , 20 3 Note that there are two other possibilities trying to solve algebraically 1 2 𝑥−6=2𝑥−4 and 1 2 𝑥−6=−(2𝑥−4) Neither of these gives a correct solution – look at the graph

Put the negative sign on the ‘easiest’ side 𝑥 2 −6𝑥+8=0 WB6 a) sketch the graph of 𝑦= 𝑥 2 −6𝑥 b) hence, solve the equation 𝑥 2 −6𝑥 =8 b) Four intersections 1st 𝑥 2 −6𝑥=8 𝑥 2 −6𝑥−8=0 𝑥= 6± 36+32 2 = 𝟑± 𝟏𝟕 2nd 𝑥 2 −6𝑥=−8 Put the negative sign on the ‘easiest’ side 𝑥 2 −6𝑥+8=0 (𝑥−4)(𝑥−2)=0 𝒙=𝟐 𝒙=𝟒

WB7 𝑓 𝑥 = 2𝑥−6 , 𝑥 𝜖 𝑅 a) Sketch the graph with equation 𝑦=𝑓 𝑥 showing the coordinates of the points where the graph cuts or meets the axes b) solve 𝑓 𝑥 =10+𝑥   𝑔 𝑥 = 𝑥 2 −2𝑥+4, 𝑥 𝜖 𝑅, 0≤𝑥<7 find 𝑓𝑔 5 Find the range of 𝑔 𝑥 2𝑥−6 =10+𝑥 2𝑥−6=10+𝑥 gives 𝑥=16 2𝑥−6=−(10+𝑥) gives 𝑥=− 4 3 𝑐) 𝑓𝑔 5 =𝑓 19 = 2(19)−6 =32 𝑑) 𝑔 𝑥 = (𝑥−1) 2 + 3, 𝑔 𝑥 ≥3 𝑔 0 =4 , 𝑔 7 =39 range is 3≤ 𝑔 𝑥 ≤39

Hence there is a root in the interval [-4, -3] WB8 𝑓 𝑥 = 𝑥 4 −4𝑥−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 𝑥 =(𝑥−4)( 𝑥 3 +𝑎 𝑥 2 +𝑏𝑥+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 −4 =32 𝑓 −3 =−147 Hence there is a root in the interval [-4, -3] 𝑓 ′ (𝑥)=4 𝑥 3 −4 𝑓 ′ (𝑥)=0 when 𝑥=1 gives point 1, −243 c) 𝑓(𝑥)=(𝑥−4)( 𝑥 3 +4 𝑥 2 +16𝑥+60)

g is defined by: f x = 𝑥 2 −4𝑥+11, 𝑥≥0 8 marks 12 mins WB9 Exam Q f is defined by: f x = 𝑥−4 −3, 𝑥∈𝑅 g is defined by: f x = 𝑥 2 −4𝑥+11, 𝑥≥0 Solve f x =3 State the range of g(x) find g𝐟 −𝟐 plus a) 𝒙−𝟒 −𝟑=𝟑 𝑥−4=6, 𝑥=10 𝒙−𝟒 =6 minus 𝑥−4=−6, 𝑥=−2 b ) 𝒙 𝟐 −𝟒𝒙+𝟏𝟏= 𝒙−𝟐 𝟐 +𝟕 Range is g(x)≥𝟕 𝐟 −𝟐 = −𝟔 −𝟑 =𝟑 g𝐟 −𝟐 = g 𝟑 = (𝟑) 𝟐 −𝟒 𝟑 +𝟏𝟏=𝟖

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 –

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