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C2 Chapter 9 Differentiation Dr J Frost (email@example.com) Last modified: 5 th October 2013
x1x1 x2x2 f(x 1 ) f(x 2 ) A function is increasing if for any two values of x, x 1 and x 2 where x 2 > x 1, then f(x 2 ) ≥ f(x 1 ) A function is strictly increasing if f(x 2 ) > f(x 1 ) 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. Increasing Functions ?
Show that f(x) = x 3 + 24x + 3 (x ϵ ℝ ) is an increasing function. Showing a function is increasing/decreasing ?
ab This is a decreasing function in the interval (a,b) i.e. where a < x < b Increasing/Decreasing in an Interval Find the values of x for which the function f(x) = x 3 + 3x 2 – 9x is a decreasing function. 3x 2 + 6x – 9 < 0 Thus -3 < x < 1 ? Find the values of x for which the function f(x) = x + (25/x) is a decreasing function. 1 – (25/x 2 ) < 0 Thus -5 < x < 5 ? 1 2
Features you’ve previously used to sketch graphs? f’(x) = 0 Stationary points are those for which f’(x) = 0 Maximum pointMinimum point Maximum/minimum points are known as ‘turning points’. Stationary Points
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... ?
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. Do we have a minimum or maximum point? ??
Find the coordinates of the turning point on the curve with equation y = x 4 – 32x. Determine whether this is a minimum or maximum point. (2, -48) Method 1Method 2 Value of x Gradient Shape x < 2 e.g. x = 1.9 x = 2 x > 2 e.g. x = 2.1 e.g. -4.56 e.g. 5.040 We can see from this shape that this is a minimum point. What are the advantages of each method? Do we have a minimum or maximum point? ??? ??? ? ? ?
Points of inflection A point of inflection is a point where the curve changes from concave to convex (or vice versa). Stationary point of inflection (“saddle point”) Non-stationary point of inflection
Stationary Points of inflection So how can we tell if a stationary point is a point of inflection?
Non-Stationary Points of inflection (not in the A Level syllabus) At this point: > 0 (i.e. not stationary) = 0 (i.e. gradient is not changing at this point) ? ?
Stationary Point Summary d 2 y / dx 2 Type of Point < 0Maximum > 0Minimum = 0Could be maximum, minimum or point of inflection. Use ‘Method 1’ to find gradient just before and after. y = x 4 has a turning point at x = 0. Show that this is a minimum point. dy/dx = 4x 3. d 2 y/dx 2 = 12x 2 When x = 0, d 2 y/dx 2 = 0, so we can’t classify immediately. When x = -0.1, dy/dx = -0.004. When x = +0.1, dy/dx = +0.004. Gradient goes from negative to positive, so minimum point. ?
Further Examples Find the stationary points of y = 2x 3 – 15x 2 + 24x + 6 and determine which of the points are maximum/minimum/points of inflection. State the range of outputs of 6x – x 2 (1, 17) is a maximum point. (4, -10) is a maximum point. ? ?
Differentiation – Practical applications Dr Frost Objectives: Use differentiation in real-life problems that involve optimisation of some variable.
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 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? x y 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? Optimisation Problems
r cm N M O 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? 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) Strategy 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. Breaking down optimisation problems
r cm N M O 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? a) Show that A = 50r – r 2 Given that r varies, find: b) The value of r for which A is a maximum and show that A is a maximum. c) Find the value of angle MON for this maximum area. d) Find the maximum area of the sector OMN. Example 1