Download presentation

1
**C2 Chapter 9 Differentiation**

Dr J Frost Last modified: 5th October 2013

2
Increasing Functions How could we use differentiation to tell us if this is a strictly increasing function? ? ...if the gradient is always positive, i.e. f’(x) > 0 for all x. x1 x2 f(x1) f(x2) A function is increasing if for any two values of x, x1 and x2 where x2 > x1, then f(x2) ≥ f(x1) A function is strictly increasing if f(x2) > f(x1)

3
**Example Exam Question 𝑑𝑦 𝑑𝑥 =2𝑥− 1 2 𝑘 𝑥 − 1 2 ? 𝑘>32 ?**

Edexcel C2 June 2010 𝑑𝑦 𝑑𝑥 =2𝑥− 1 2 𝑘 𝑥 − 1 2 𝑘>32 ? ?

4
**Show that f(x) = x3 + 24x + 3 (x ϵ ℝ) is an increasing function.**

Showing a function is increasing/decreasing Show that f(x) = x3 + 24x + 3 (x ϵ ℝ) is an increasing function. ? 𝒇 ′ 𝒙 =𝟑 𝒙 𝟐 +𝟐𝟒 𝟑 𝒙 𝟐 is always positive for all 𝒙. Thus 𝒇 ′ 𝒙 >𝟎

5
**Increasing/Decreasing in an Interval**

This is a decreasing function in the interval (a,b) i.e. where a < x < b a b 1 Find the values of x for which the function f(x) = x3 + 3x2 – 9x is a decreasing function. ? 3x2 + 6x – 9 < 0 Thus -3 < x < 1 2 Find the values of x for which the function f(x) = x + (25/x) is a decreasing function. ? 1 – (25/x2) < 0 Thus -5 < x < 5

6
Questions C2 pg 130 Exercise 9A

7
**Stationary Points Features you’ve previously used to sketch graphs?**

f’(x) = 0 Maximum point Stationary points are those for which f’(x) = 0 Minimum point f’(x) = 0 Maximum/minimum points are known as ‘turning points’.

8
**Finding turning points**

Edexcel C2 May 2013 (Retracted) ? (2,9) Although it might be interest to know if this is a minimum point or a maximum point...

9
**Do we have a minimum or maximum point?**

A maximum point is often called a ‘peak’ and a minimum point a ‘trough’. Method 1: Consider the points immediately before and after the stationary point. Method 2: Use the second-order derivative to see whether the gradient is increasing or decreasing. ? ?

10
**Do we have a minimum or maximum point?**

(2, -48) Find the coordinates of the turning point on the curve with equation y = x4 – 32x. Determine whether this is a minimum or maximum point. Method 1 Method 2 𝑑 2 𝑦 𝑑 𝑥 2 = 12x2 ? x < 2 e.g. x = 1.9 x > 2 e.g. x = 2.1 x = 2 Value of x When x = 2, 𝑑 2 𝑦 𝑑 𝑥 2 = 48. ? ? ? ? Gradient e.g e.g. 5.04 ? ? ? Shape We can see from this shape that this is a minimum point. 𝑑 2 𝑦 𝑑 𝑥 2 > 0, so a minimum point. ? What are the advantages of each method?

11
Points of inflection Stationary point of inflection (“saddle point”) Non-stationary point of inflection A point of inflection is a point where the curve changes from concave to convex (or vice versa). We can see that when 𝑑𝑦 𝑑𝑥 =0, we might not have a maximum or minimum, but a point of inflection instead. At A Level, you won’t see non-stationary points of inflection.

12
**Stationary Points of inflection**

So how can we tell if a stationary point is a point of inflection?

13
**Non-Stationary Points of inflection**

(not in the A Level syllabus) At this point: 𝑑𝑦 𝑑𝑥 ? > 0 (i.e. not stationary) 𝑑 2 𝑦 𝑑 𝑥 2 ? = 0 (i.e. gradient is not changing at this point)

14
**Stationary Point Summary**

d2y / dx2 Type of Point < 0 Maximum > 0 Minimum = 0 Could be maximum, minimum or point of inflection. Use ‘Method 1’ to find gradient just before and after. y = x4 has a turning point at x = 0. Show that this is a minimum point. ? dy/dx = 4x3. d2y/dx2 = 12x2 When x = 0, d2y/dx2 = 0, so we can’t classify immediately. When x = -0.1, dy/dx = When x = +0.1, dy/dx = Gradient goes from negative to positive, so minimum point.

15
Further Examples Find the stationary points of y = 2x3 – 15x2 + 24x + 6 and determine which of the points are maximum/minimum/points of inflection. ? (1, 17) is a maximum point. (4, -10) is a maximum point. State the range of outputs of 6x – x2 ? 𝑓 𝑥 ≤9

16
Exercise 9B

17
**Differentiation – Practical applications**

Dr Frost Differentiation – Practical applications Objectives: Use differentiation in real-life problems that involve optimisation of some variable.

18
**Optimisation Problems**

We have a sheet of A4 paper, which we want to fold into a cuboid. What height should we choose for the cuboid to maximise the volume? These are examples of optimisation problems: we’re trying to maximise/minimise some quantity by choosing an appropriate value of a variable that we can control. We have 50m of fencing, and want to make a bear pen of the following shape, but that maximises the area. What should we choose x and y to be? x y

19
**Breaking down optimisation problems**

Suppose that we have 100cm of rope, that we want to put in the shape of a minor segment. We want to choose a radius for this minor segment that maximises the area covered by the rope. What radius do we choose? M N r cm Strategy 1. Form two equations: one representing the thing we’re trying to maximise (here the area) and the other representing the constraint (here the length of rope) O 2. Use calculus to find out the value of the variable we’re interested in when we have a minimum/maximum. e.g. Find when 𝑑𝐴 𝑑𝑟 =0. Typically we’d need to write out two equations (e.g. perimeter and area, or volume and area) and combine them together, using given information, to form the one equation we’d need.

20
Example Exam Question Edexcel C2 May 2011

21
Example 1 Suppose that we have 100cm of rope, that we want to put in the shape of a minor segment. We want to choose a radius for this minor segment that maximises the area covered by the rope. What radius do we choose? M N a) Show that A = 50r – r2 r cm Given that r varies, find: b) The value of r for which A is a maximum and show that A is a maximum. O c) Find the value of angle MON for this maximum area. d) Find the maximum area of the sector OMN.

22
Example 2 A large tank in the shape of a cuboid is to be made from 54m2 of sheet metal. The tank has a horizontal base and no top. The height of the tank is 𝑥 metres. Two of the opposite vertical faces are squares. a) Show that the volume, V m3, of the tank is given by 𝑽=𝟏𝟖𝒙− 𝟐 𝟑 𝒙 𝟑 . b) Given that x can vary, use differentiation to find the maximum or minimum value of V. c) Justify whether your value of V is a minimum or maximum. 𝒙 𝒙 𝒚

Similar presentations

OK

DERIVATIVES 3. Summary f(x) ≈ f(a) + f’(a)(x – a) L(x) = f(a) + f’(a)(x – a) ∆y = f(x + ∆x) – f(x) dx = ∆x dy = f’(x)dx ∆ y≈ dy.

DERIVATIVES 3. Summary f(x) ≈ f(a) + f’(a)(x – a) L(x) = f(a) + f’(a)(x – a) ∆y = f(x + ∆x) – f(x) dx = ∆x dy = f’(x)dx ∆ y≈ dy.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Viewer ppt online Ppt on video teleconferencing services Ppt on automobile industry in india 2012 Ppt on trial and error quotes Convert free pdf to ppt online free Ppt on inhabiting other planets like earth Ppt on articles of association pdf Ppt on panel discussion questions Ppt on the elements of the short story Ppt on geotextile