Increasing and decreasing Functions

Increasing and decreasing KUS objectives BAT understand increasing and decreasing parts of functions Starter: Differentiate 𝑦= 𝑥 3 −2 𝑥 2 +𝑥−3 Find the gradient of the curve 𝑦= 𝑥 3 −2 𝑥 2 +𝑥−3 at the point (2, -1) Find the equation of the tangent to the curve 𝑦= 𝑥 3 −2 𝑥 2 +𝑥−3 at the point (2, -1)

y You need to know the difference between Increasing and Decreasing Functions An increasing function is one with a positive gradient. A decreasing function is one with a negative gradient. This function is increasing for all values of x x y This function is decreasing for all values of x x

𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 2 −𝑥−6 Point (0, −6) 𝑑𝑦 𝑑𝑥 =(𝑥−3)(𝑥+2) WB1 𝑓 𝑥 = 1 3 𝑥 3 − 1 2 𝑥 2 −6𝑥+2 Draw the gradient graph of 𝑓(𝑥) showing clearly any intersections with the axes and the stationary point Explain these features of the gradient graph, giving references to the graph of 𝑓(𝑥) i) Intersections of the gradient graph with the x axis ii) The minimum point on the gradient graph 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 2 −𝑥−6 Point (0, −6) 𝑑𝑦 𝑑𝑥 =(𝑥−3)(𝑥+2) Points (3, 0) and (−2, 0) 𝑑 2 𝑦 𝑑 𝑥 2 =2𝑥−1 =0 when 𝑥= 1 2 , Point 1 2 , − 13 12 Gradient graph b) i) Intersections with the x axis are where the gradient is zero (stationary points) ii) The minimum point on the gradient graph is the steepest negative gradient on f(x) – where the rate of change is its lowest negative value

𝑑𝑦 𝑑𝑥 = 5+3𝑥−2𝑥 2 Point (0, 5) 𝑑𝑦 𝑑𝑥 =(𝑥+1)(5−2𝑥) WB2 𝑓 𝑥 =5𝑥+ 3 2 𝑥 2 − 2 3 𝑥 3 Draw the gradient graph of 𝑓(𝑥) showing clearly any intersections with the axes and the stationary point find the maximum value of 𝑓′ 𝑥 and the gradient and y value at that point 𝑑𝑦 𝑑𝑥 = 5+3𝑥−2𝑥 2 Point (0, 5) 𝑑𝑦 𝑑𝑥 =(𝑥+1)(5−2𝑥) Points (−1, 0) and 5 2 , 0 𝑑 2 𝑦 𝑑 𝑥 2 =3−4𝑥 =0 when 𝑥= 3 4 , Point 3 4 , 69 16 on f(x) when 𝑥= 3 4 , 𝑑𝑦 𝑑𝑥 = 49 8 𝑑𝑦 𝑑𝑥 Gradient graph

𝑑𝑦 𝑑𝑥 = 𝑥 3 −3𝑥=0 when 𝑥=−1 and 𝑥=1 𝑓 𝑥 -1 1 WB3 𝑓 𝑥 = 𝑥 3 −3𝑥 Find 𝑓′ 𝑥 find the values of 𝑥 where the gradient is negative, express your answer in the form 𝑎<𝑥<𝑏 where 𝑎 and 𝑏 are integers f(x) 𝑑𝑦 𝑑𝑥 = 3𝑥 2 −3 𝑑𝑦 𝑑𝑥 = 𝑥 3 −3𝑥=0 when 𝑥=−1 and 𝑥=1 𝑓 𝑥 -1 1 𝒇 ′ 𝒙 + ve - ve + ve So the gradient is negative when −1<𝑥<1

So the gradient is positive WB4 𝑓 𝑥 = 𝑥 4 +2 𝑥 3 −3 𝑥 2 Find 𝑓′ 𝑥 find the values of 𝑥 where the gradient is positive, express your answers in the form 𝑎<𝑥<𝑏 𝑑𝑦 𝑑𝑥 =4 𝑥 3 +6 𝑥 2 −6𝑥 =2𝑥( 2𝑥 2 +3𝑥−6) f(x) 𝑑𝑦 𝑑𝑥 = =0 when 𝑥=0 and 𝑥= −3± 57 4 𝑓 𝑥 −3− 57 4 −3+ 57 4 𝒇 ′ 𝒙 - ve + ve - ve + ve So the gradient is positive when −3− 57 4 <𝑥<0 and 𝑥> −3+ 57 4

Decreasing Function range WB5 Find the range of values where: 𝑓 𝑥 = 𝑥 3 +3 𝑥 2 −9𝑥 Is a decreasing function y f(x) Differentiate for the gradient function We want the gradient to be below 0 Factorise x Factorise again Normally x = -3 and 1 BUT, we want values that will make the function negative… -3 1 Decreasing Function range

WB6 Show that the function 𝑓 𝑥 = 𝑥 3 +24𝑥+3 is an increasing function. f(x) Differentiate to get the gradient function  Since x2 has to be positive, 3x2 + 24 will be as well  So the gradient will always be positive, hence an increasing function

Differentiate to get the gradient function Show that 𝑦=2− 𝑥 2 − 1 𝑥 2 is a decreasing function when −1<𝑥<0 and 𝑥>1 WB7 f(x) 𝑦=2− 𝑥 2 − 𝑥 −2 Differentiate to get the gradient function 𝑑𝑦 𝑑𝑥 =−2𝑥+2 𝑥 −3 2−2 𝑥 4 𝑥 3 = 2(1− 𝑥 4 ) 𝑥 3 𝑑𝑦 𝑑𝑥 =−2𝑥+ 2 𝑥 3 = 𝑑𝑦 𝑑𝑥 =0 when 𝑥=±1 but also there is an asymptote at 𝑥=0 𝑓 𝑥 -1 1 𝒇 ′ 𝒙 + ve - ve + ve - ve So y is decreasing when −1<𝑥<0 and 𝑥>1

𝑑𝑦 𝑑𝑥 =3𝑥−𝑐 𝑑𝑦 𝑑𝑥 >0 when 𝑥 > 𝑐 3 𝑑𝑦 𝑑𝑥 =3𝑥−𝑐 Gradient graph 𝑐 3 WB8 Show that the function 𝑓 𝑥 = 3 2 𝑥 2 −𝑐𝑥+7 is an increasing function when 𝑥> 𝑐 3 𝑑𝑦 𝑑𝑥 =3𝑥−𝑐 Gradient graph 𝑐 3 𝑑𝑦 𝑑𝑥 =3𝑥−𝑐 𝑑𝑦 𝑑𝑥 >0 when 𝑥 > 𝑐 3

solving gives 𝑡=2± 3 the answer that fits is 𝑡=2− 3 =0.268 𝑠𝑒𝑐𝑠 WB9a A graph for the motion of a firework in the first two second of its journey is shown. the velocity of the firework’s journey Is modelled by the equation 𝑡𝑖𝑚𝑒 (𝑠𝑒𝑐𝑠) 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣 𝑚𝑠 −1 v=10 8+ 𝑡−2 3 a) find the time when the velocity of the firework reaches 40 ms-1 b) find the time when the acceleration of the firework reaches 90 ms-2 c) give a reason why the model does not accurately predict the motion of the firework after 2 seconds 𝑎) 𝑣=10 8+ 𝑡−2 3 =40 𝑡−2 3 =−4 𝑡= 3 −4 +2=0.413 𝑠𝑒𝑐𝑠 𝑏) 𝑣=10 𝑡 3 −60 𝑡 2 +120𝑡+72 𝑑𝑣 𝑑𝑡 =30 𝑡 2 −120𝑡+120=30 ( 𝑡 2 −4𝑡+4) =90 solving gives 𝑡=2± 3 the answer that fits is 𝑡=2− 3 =0.268 𝑠𝑒𝑐𝑠 𝑐) after 2 secs the firework has used up its fuel / is no longer being pushed upwards

the answer that fits is 𝑡=2− 3 =0.268 𝑠𝑒𝑐𝑠 WB9b A graph for the motion of a firework in the first two second of its journey is shown. the velocity of the firework’s journey Is modelled by the equation v=10 8+ 𝑡−2 3 a) find the time when the velocity of the firework reaches 40 ms-1 b) find the time when the acceleration of the firework reaches 90 ms-2 c) give a reason why the model does not accurately predict the motion of the firework after 2 seconds 𝑡𝑖𝑚𝑒 (𝑠𝑒𝑐𝑠) 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣 𝑚𝑠 −1 𝑎) 𝑣=10 8+ 𝑡−2 3 =40 𝑡−2 3 =−4 𝑡= 3 −4 +2=0.413 𝑠𝑒𝑐𝑠 𝑏) 𝑣=10 𝑡 3 −60 𝑡 2 +120𝑡+72 𝑑𝑣 𝑑𝑡 =30 𝑡 2 −120𝑡+120=30 ( 𝑡 2 −4𝑡+4) =90 solving gives 𝑡=2± 3 the answer that fits is 𝑡=2− 3 =0.268 𝑠𝑒𝑐𝑠 𝑐) after 2 secs the firework has used up its fuel / is no longer being pushed upwards

WB10a A graph for the horizontal motion of a pendulum is shown WB10a A graph for the horizontal motion of a pendulum is shown. the velocity of the pendulum is modelled by the equation v(t)=3+ 2cos 2𝑡 a) find the time (after 𝑡=0) when the velocity of the pendulum first reaches 4 ms-1 b) the gradient function of the graph is 𝑑𝑣 𝑑𝑡 =−4 sin 2𝑡 find the time (after 𝑡=0) when the acceleration is zero c) explain what the pendulum is doing when the acceleration is zero 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣 𝑚𝑠 −1 𝑡𝑖𝑚𝑒

So the first value of t is 𝑡=30 WB10b A graph for the horizontal motion of a pendulum is shown. the velocity of the pendulum is modelled by the equation v(t)=3+ 2cos 2𝑡 a) find the time (after 𝑡=0) when the velocity of the pendulum first reaches 4 ms-1 b) the gradient function of the graph is 𝑑𝑣 𝑑𝑡 =−4 sin 2𝑡 find the time (after 𝑡=0) when the acceleration is zero c) explain what the pendulum is doing when the acceleration is zero 𝑎) v=3+ 2cos 2𝑡 =4 cos 2𝑡 = 1 2 2𝑡=…, 60, 300, …. So the first value of t is 𝑡=30 b) Acceleration is the rate of change of velocity 𝑑𝑣 𝑑𝑡 =−4 sin 2𝑡 Acceleration is zero when −4 sin 2𝑡 =0 2𝑡=…, 0, 180, 369, … So the first value of t is 𝑡=90

One thing to improve is – KUS objectives BAT understand increasing and decreasing parts of functions self-assess One thing learned is – One thing to improve is –

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