1. Increasing and decreasing
Functions

2. 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)

3. 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

4. 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 2 −𝑥−6 Point (0, −6) 𝑑𝑦 𝑑𝑥 =(𝑥−3)(𝑥+2)
WB 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 , − 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

5. 𝑑𝑦 𝑑𝑥 = 5+3𝑥−2𝑥 2 Point (0, 5) 𝑑𝑦 𝑑𝑥 =(𝑥+1)(5−2𝑥)
WB 𝑓 𝑥 =5𝑥 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 , 0
𝑑 2 𝑦 𝑑 𝑥 2 =3−4𝑥
=0 when 𝑥= 3 4 , Point , on f(x)
when 𝑥= 3 4 , 𝑑𝑦 𝑑𝑥 = 49 8
𝑑𝑦 𝑑𝑥
Gradient graph

6. 𝑑𝑦 𝑑𝑥 = 𝑥 3 −3𝑥=0 when 𝑥=−1 and 𝑥=1 𝑓 𝑥 -1 1
WB 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

7. So the gradient is positive
WB 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±
𝑓 𝑥
−3− −
𝒇 ′ 𝒙
- ve
+ ve
- ve
+ ve
So the gradient is positive
when −3− <𝑥<0 and 𝑥> −

8. 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

9. 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, 3x will be as well
 So the gradient will always be positive, hence an increasing function

10. 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 = 2(1− 𝑥 4 ) 𝑥 3
𝑑𝑦 𝑑𝑥 =−2𝑥+ 2 𝑥 =
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

11. 𝑑𝑦 𝑑𝑥 =3𝑥−𝑐 𝑑𝑦 𝑑𝑥 >0 when 𝑥 > 𝑐 3 𝑑𝑦 𝑑𝑥 =3𝑥−𝑐 Gradient graph 𝑐 3
WB 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

12. 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= 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= 𝑡−2 3
=40
𝑡−2 3 =−4
𝑡= 3 −4 +2=0.413 𝑠𝑒𝑐𝑠
𝑏) 𝑣=10 𝑡 3 −60 𝑡 𝑡+72
𝑑𝑣 𝑑𝑡 =30 𝑡 2 −120𝑡+120=30 ( 𝑡 2 −4𝑡+4)
=90
solving gives 𝑡=2± 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

13. 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= 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= 𝑡−2 3
=40
𝑡−2 3 =−4
𝑡= 3 −4 +2=0.413 𝑠𝑒𝑐𝑠
𝑏) 𝑣=10 𝑡 3 −60 𝑡 𝑡+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

14. 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

15. 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

16. 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 –

17. END