1. Tangents and Rates of Change – Outcomes
Interpret the derivative as a gradient function.
Interpret the derivative as a rate of change.
Solve problems about tangents and normals to curves.
Solve problems about maxima and minima.
Solve problems about points of inflexion.
Solve kinematic problems.
Solve optimization problems.
Discuss the behaviour of graphs and the relationship between the graphs of 𝑓(𝑥), 𝑓′(𝑥), and 𝑓′′(𝑥).

2. Interpret Derivative as Gradient
The notation d𝑦 d𝑥 loosely comes from the slope formula:
𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 or more simply 𝑚= Δ𝑦 Δ𝑥 .
In fact, derivatives in general represent the slope (or gradient) of a function.
e.g. Find the derivative of 𝑦=5𝑥−3, then find its slope by generating points.
d𝑦 d𝑥 =5
Using (0, −3) and (1,2)
𝑚= 2−−3 1−0 = 5 1 =5

3. Interpret Derivative as Gradient
For functions that do not have a constant slope (any polynomial of order ≠1, exponentials, logs, trig functions…), the derivative represents the slope of the tangent to that function.
𝑦=− 𝑥 2 +7𝑥−10
⇒ d𝑦 d𝑥 =−2𝑥+7

4. Interpret Derivative as Rate of Change
Slopes are a useful visual representation for curves, but they also represent how quickly one variable changes with the other.
e.g. For a line with slope 5, for every +1 change in 𝑥, there is a +5 change in 𝑦.
More generally, the slope can be interpreted as the rate of change of a function.
For any non-linear function where the slope and rate of change are not constant, the derivative represents an instantaneous rate of change at a particular location on the function.

5. Solve Problems about Tangents
e.g. Find the value of the slope / rate of change at 𝑥=2 for each of the following functions:
𝑓 𝑥 =3 𝑥 2 +5𝑥
𝑓 𝑥 =2 𝑥 2 +12𝑥+3
𝑓 𝑥 =1−𝑥+ 𝑥 2
𝑓 𝑥 =6𝑥+10
𝑓 𝑥 = 𝑥 3 −7 𝑥 2 +6𝑥
𝑓 𝑥 = 𝑥 2 −5𝑥+6 𝑥−3
𝑓 𝑥 =5 𝑥 2 −8𝑥+ 𝑧 2

6. Solve Problems about Normals
Recall that a normal and a tangent are perpendicular.
Recall a result from coordinate geometry for perpendicular lines:
𝑚 1 ⊥ 𝑚 2 ⇒ 𝑚 1 =− 1 𝑚 2 or 𝑚 1 × 𝑚 2 =−1
e.g. Find the equation to the normal of the function 𝑓 𝑥 = 𝑥 2 −6𝑥+9 at 𝑥=2
𝑚 𝑇 = 𝑓 ′ 𝑥 =2𝑥−6
𝑚 𝑇 2 = 𝑓 ′ 2 =2 2 −6=−2
𝑚 𝑁 =− 1 𝑚 𝑇 =− 1 −2 = 1 2

7. Solve Problems about Normals
Need the point of contact to use equation of a line formula:
𝑓 2 = 2 2 −6 2 +9=4−12+9=1
𝑦− 𝑦 1 = 𝑚 𝑁 𝑥− 𝑥 1
⇒𝑦−1= 1 2 𝑥−2
⇒2𝑦−2=𝑥−2
⇒2𝑦=𝑥
⇒𝑦= 1 2 𝑥

8. Solve Problems about Normals
Find the equation of the normal at 𝑥=2 for each of the following curves:
𝑓 𝑥 =3 𝑥 2 +5𝑥
𝑓 𝑥 =2 𝑥 2 +12𝑥+3
𝑓 𝑥 =1−𝑥+ 𝑥 2
𝑓 𝑥 =6𝑥+10
𝑓 𝑥 = 𝑥 3 −7 𝑥 2 +6𝑥
𝑓 𝑥 = 𝑥 2 −5𝑥+6 𝑥−3
𝑓 𝑥 =5 𝑥 2 −8𝑥

9. Solve Problems about Maxima and Minima
Important points for curves are turning points / stationary points / extrema.
They come in two types: maximum and minimum.
We often refer to “local” maximum and minimum to distinguish when there are more than one of either

10. Solve Problems about Maxima and Minima
If the derivative of a function represents the slope of the tangent, what value does the derivative take at turning points?

11. Solve Problems about Maxima and Minima
e.g. Find the coordinates of the maximum and minimum points of 𝑓 𝑥 = 𝑥 3 −9 𝑥 2 +24𝑥−10
For turning points, d𝑦 d𝑥 =0
⇒3 𝑥 2 −18𝑥+24=0
⇒ 𝑥 2 −6𝑥+8=0
⇒ 𝑥−4 𝑥−2 =0
⇒𝑥=4
𝑓 4 = 4 3 − −10 =6
4,6 - minimum
⇒𝑥=2
𝑓 2 = 2 3 − −10
=10
2,10 - maximum
How do you know which is maximum and minimum?

12. Solve Problems about Maxima and Minima
Know your curve:
𝑥 3 −9 𝑥 2 +24𝑥−10 is a positive cubic so looks like
You can use 𝑥-coordinates to decide where the various maxima and minima are.

13. Solve Problems about Maxima and Minima
Alternatively, use the second derivative test:
For minima, d 2 𝑦 d 𝑥 2 >0
For maxima, d 2 𝑦 d 𝑥 2 <0
d 2 𝑦 d 𝑥 2 =6𝑥−18 for this function.
d 2 𝑦 d 𝑥 =6 4 −18=6>0⇒ minimum
d 2 𝑦 d 𝑥 =6 2 −18=−6<0⇒ maximum

14. Solve Problems about Maxima and Minima
It may also be important to note where a function is increasing or decreasing.
This happens everywhere the function is defined except for maxima and minima.
Increasing: d𝑦 d𝑥 >0
Decreasing: d𝑦 d𝑥 <0
Alternatively, just look at the graph:

15. Solve Problems about Maxima and Minima
e.g. Find the coordinates of each of the extrema of the following and hence note where they are increasing and decreasing:
𝑓 𝑥 = 𝑥 2 −5𝑥+6
𝑓 𝑥 = 𝑥 2 +2𝑥−15
𝑦= 𝑥 3 −5 𝑥 2 +7𝑥−5
𝑦= 𝑥 3 −6 𝑥 2 +12𝑥+1
𝑓 𝑥 = 𝑥 4 − 𝑥 2
𝑓 𝑥 =𝑥 𝑒 𝑥
𝑔 𝑥 = ln 𝑥 𝑥

16. Solve Problems about Maxima and Minima
QB T6P3 Q15
Let 𝑓 𝑥 = 𝑥 −2 𝑥 2 +5𝑥−2 for −2≤𝑥≤4, 𝑥≠ 1 2 , 𝑥≠2. The graph of 𝑓 is shown.
The graph of 𝑓 has a local minimum at 𝐴 1,1 and a local maximum at 𝐵.
Use the quotient rule to show that 𝑓 ′ 𝑥 = 2 𝑥 2 −2 −2 𝑥 2 +5𝑥−
Hence find the coordinates of 𝐵.
Given that the line 𝑦=𝑘 does not meet the graph of 𝑓, find the possible values of 𝑘.

17. Solve Problems about Maxima and Minima
QB T6P1 Q27
Consider 𝑓 𝑥 = 𝑥 2 + 𝑝 𝑥 , 𝑥≠0, where 𝑝 is a constant.
Find 𝑓 ′ (𝑥).
There is a minimum value of 𝑓(𝑥) when 𝑥=−2. Find the value of 𝑝.

18. Solve Problems about Inflexion
A point of inflexion is like a meta-extremum.
For extrema, d𝑦 d𝑥 =0, while
For points of inflexion, d 2 𝑦 d 𝑥 2 =0.
Graphically these are where curvature changes direction.
e.g. in this cubic, it changes from concave downward (∩) to concave upward (∪) as 𝑥 increases.

19. Solve Problems about Inflexion
Find the 𝑥-coordinate(s) of the point(s) of inflexion for each of the following functions:
𝑦= 𝑥 3 + 𝑥 2 −8𝑥+5
𝑦= 𝑥 3 −5 𝑥 2 +7𝑥−5
𝑦= 𝑥 3 −6 𝑥 2 +12𝑥+1
𝑦= 𝑥 𝑥 2 +1
𝑦= 𝑥 4
𝑦= 𝑥 2 −5𝑥+6

20. Solve Problems about Points of Inflexion
QB T6P1 Q8
The following diagram shows the graph of 𝑓 𝑥 = 𝑒 − 𝑥 2 .
The points 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 lie on the graph of 𝑓. Two of these are points of inflexion.
Identify the two points of inflexion.
Find 𝑓 ′ 𝑥
Show that 𝑓 ′′ 𝑥 = 4 𝑥 2 −2 𝑒 − 𝑥 2 .
Find the 𝑥-coordinate of each point of inflexion.
Use the second derivative to show that one of these points is a point of inflexion.

21. Solve Problems about Points of Inflexion
QB T6P1 Q22
Let 𝑓 ′ 𝑥 =−24 𝑥 3 +9 𝑥 2 +3𝑥+1
There are two points of inflexion on the graph 𝑓. Write down the 𝑥-coordinates of these points.
Let 𝑔 𝑥 = 𝑓 ′′ 𝑥 . Explain why the graph of 𝑔 has no points of inflexion.
QB T6P2 Q9
A function 𝑓 has its first derivative 𝑓 ′ 𝑥 = 𝑥−3 3 .
Find the second derivative.
Find 𝑓′(3) and 𝑓′′(3).
The point 𝑃 on the graph of 𝑓 has 𝑥-coordinate 3. Explain why 𝑃 is not a point of inflexion.

22. Discuss 𝑓 𝑥 , 𝑓 ′ 𝑥 , and 𝑓 ′′ 𝑥
Here is a graph of a cubic and its derivatives:
What is happening to the order of the polynomial as we differentiate?
Note how the extrema of one graph corresponds to a root of its derivative

23. Discuss 𝑓 𝑥 , 𝑓 ′ 𝑥 , and 𝑓 ′′ 𝑥
Here is a graph of 𝑒 2𝑥 and its derivatives:
Note how similar the graphs are as they are just a factor of 2 apart
Note one way to tell them apart – the 𝑦-intercepts

24. Discuss 𝑓 𝑥 , 𝑓 ′ 𝑥 , and 𝑓 ′′ 𝑥
Here is a graph of ln 𝑥 and its derivatives:
Note that the derivatives are asymptotic with the axes due to the division by 𝑥
Note the difference between 𝑓 ′ 𝑥 and 𝑓 ′′ 𝑥 around 𝑥=−1

25. Discuss 𝑓 𝑥 , 𝑓 ′ 𝑥 , and 𝑓 ′′ 𝑥
QB T6P2 Q1
The diagram shows part of the graph of 𝑦=𝑓′(𝑥). The 𝑥-intercepts are at points 𝐴 and 𝐶. There is a minimum at 𝐵, and a maximum at 𝐷.
Write down the value of 𝑓′(𝑥) at 𝐶.
Hence, show that 𝐶 corresponds to a minimum on the graph of 𝑓, i.e. that it has the same 𝑥-coordinate.
Which of the points 𝐴, 𝐵, 𝐷 corresponds to a maximum on the graph of 𝑓?
Show that 𝐵 corresponds to a point of inflexion on the graph of 𝑓.

26. Solve Optimization Problems
In context problems, extrema are referred to by appropriate adjectives:
Maxima: most, largest, biggest, greatest, tallest, longest, heaviest, highest, fastest, quickest, hardest, best, strongest, richest, etc.
Minima: fewest, smallest, least, shortest, lightest, lowest, slowest, softest, worst, weakest, poorest, etc.
The usual process for finding stationary points works here.

27. Solve Optimization Problems
QB T6P1 Q32
A farmer wishes to create a rectangular enclosure, 𝐴𝐵𝐶𝐷, of area 525 m2, as shown below.
The fencing used for side 𝐴𝐵 costs $11 per metre. The fencing for the other three sides costs $3 per metre. The farmer creates an enclosure so that the cost is a minimum. Find this minimum cost.

28. Solve Optimization Problems
16M.1.sl.TZ2.9
Fred makes an open metal container in the shape of a cuboid, as shown in the following diagram.
The container has height 𝑥 m, width 𝑥 m and length 𝑦 m. The volume is 36 m2. Let 𝐴 𝑥 be the outside surface area of the container.
Show that 𝐴 𝑥 = 108 𝑥 +2 𝑥 2
Find 𝐴 ′ 𝑥 .
Given that the outside surface area is a minimum, find the height of the container.
Fred paints the outside of the container. A tin of paint covers a surface area of 10 m2 and costs $20. Find the total cost of the tins needed to paint the container.

29. Solve Optimization Problems
2013 HL P1 Q7
A stadium can hold people. When the ticket price is €20, the expected attendance at an event is people. A survey suggests that for every €1 reduction, from €20, in the ticket price, the expected attendance would increase by 1000 people.
If the ticket price was €18, how many people would be expected to attend?
Let 𝑥 be the ticket price, where 𝑥≤20. Write down, in terms of 𝑥, the expected attendance at such an event.
Write down a function 𝑓 that gives the expected income from the sales of tickets for such an event.
Find the price at which tickets should be sold to give the maximum expected income. Hence, find this income.

30. Solve Kinematic Problems
Kinematics is a branch of the study of motion.
We look at three quantities here:
Displacement, 𝑠, describes the position of a body over time (similar to distance).
Velocity, 𝑣= d𝑠 d𝑡 , describes the rate of change of a body’s position (similar to speed).
Acceleration, 𝑎= d𝑣 d𝑡 = d 2 𝑠 d 𝑡 2 , describes the rate of change of a body’s velocity.

31. Solve Kinematic Problems
An object’s position over 2.5 seconds is given by the function 𝑠= 𝑡 3 −4 𝑡 2 +4𝑡 with 𝑠 in metres and 𝑡 in seconds. Find:
The object’s velocity over time.
The object’s acceleration over time.
The object’s minimum velocity.
The object’s maximum velocity.
The time when the object was back at its starting position.
The time when the object was not accelerating.

32. Solve Kinematic Problems
QB T6P3 Q17
A particle moves in a straight line with velocity 𝑣=12𝑡− 2 𝑡 3 −1, for 𝑡≥0, where 𝑣 is in centimetres per second and 𝑡 is in seconds.
Find the acceleration of the particle after 2.7 seconds.
QB T6P1 Q11
A particle moves along a straight line so that its velocity, 𝑣 ms-1, at a time, 𝑡 seconds is given by 𝑣=240+20𝑡 − 10 𝑡 2 , for 0≤𝑡≤6.
Find the value of 𝑡 when the speed of the particle is greatest.
Find the acceleration of the particle when its speed is zero.

33. Solve Kinematic Problems
QB T6P1 Q35
The following diagram shows the displacement, velocity, and acceleration of a moving object as functions of time, 𝑡.
Identify which graph corresponds to which function.
Write down the value of 𝑡 when the velocity is greatest.