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**C2 Chapter 9 Differentiation**

Dr J Frost Last modified: 5th October 2013

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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)

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**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 ? ?

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**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

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**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

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Questions C2 pg 130 Exercise 9A

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**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’.

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**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...

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**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. ? ?

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**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?

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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.

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**Stationary Points of inflection**

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

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**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)

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**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.

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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

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Exercise 9B

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**Differentiation – Practical applications**

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

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**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

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**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.

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Example Exam Question Edexcel C2 May 2011

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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.

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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

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