2 The slides that follow are samples from the 55 presentations that make up the work for the A2 core modules C3 and C4.2: Inverse Functions22: Integrating the Simple Functions3: Graphs of Inverse Functions28: Integration giving Logs4: The Function31: Double Angle Formulae13: Inverse Trig Ratios15: More Transformations43: Partial Fractions16: The Modulus Function47: Solving Differential Equations18: Iteration Diagrams and Convergence51: The Vector Equation of a Line
3 Explanation of Clip-art images An important result, example or summary that students might want to note.It would be a good idea for students to check they can use their calculators correctly to get the result shown.An exercise for students to do without help.
4 2: Inverse Functions Demo version note: The students have already met the formal definition of a function and the ideas of domain and range. In the following slides we prepare to introduce the condition for an inverse function.
5 Inverse FunctionsOne-to-one and many-to-one functionsConsider the following graphsandEach value of x maps to only one value of y . . .Each value of x maps to only one value of y . . .and each y is mapped from only one x.BUT many other x values map to that y.
6 Inverse FunctionsOne-to-one and many-to-one functionsConsider the following graphsandis an example of a many-to-one functionis an example of a one-to-one function
7 3: Graphs of Inverse Functions Demo version note:By this time the students know how to find an inverse function. The graphical link between a function and its inverse has also been established and this example follows.
8 Graphs of Inverse Functions e.g. On the same axes, sketch the graph ofand its inverse.N.B!Solution:x
9 Graphs of Inverse Functions e.g. On the same axes, sketch the graph ofand its inverse.N.B!Solution:N.B. Using the translation of we can see the inverse function is
10 Graphs of Inverse Functions A bit more on domain and rangeThe previous example usedThe domain of is.Since is found by swapping x and y,the values of the domain of give the values of the range ofDomainRange
11 Graphs of Inverse Functions A bit more on domain and rangeThe previous example usedThe domain of is.Since is found by swapping x and y,the values of the domain of give the values of the range ofSimilarly, the values of the range ofgive the values of the domain of
12 4: The Function Demo version note: The exponential function has been defined and here we build on earlier work to find the inverse and its graph.
13 The FunctionMore Indices and LogsWe know that( since an index is a log )The function contains the index x, so x is a log.BUT the base of the log is e not 10, soLogs with a base e are called natural logsWe write as ( n for natural ) so,
14 The FunctionThe Inverse ofis a one-to-one function so it has an inverse function.We can sketch the inverse by reflecting in y = x.Finding the equation of the inverse function is easy!We’ve already done the 1st step ofrearranging:So,Nowswap letters:N.B. The domain is
15 13: Inverse Trig Ratios Demo version note: This presentation on inverse trig ratios revises the idea of a domain. It also reminds students that a one-to-one relationship is needed in order to find the inverse function.
16 Inverse Trig RatiosWhat domain would you use?We need to make sure that we have all the y-values without any being repeated.The domain is defined as( or, in degrees, as ).
17 15: More Transformations Demo version note:As new topics are introduced, the presentations revise earlier work. Here, the exponential function is used to show the result of combining transformations.
18 x has been replaced by 2x: More TransformationsCombined Transformationse.g. 1 Describe the transformations of that give the function Hence sketch the function.Solution:x has been replaced by 2x:so we have a stretch of s.f.parallel to thex-axis1 has then been added:so we have a translation of
19 More TransformationsWe do the sketch in 2 stages:The point on the y-axis . . .doesn’t move with a stretch parallel to the x-axis
20 More TransformationsWe do the sketch in 2 stages:
21 16: The Modulus Function Demo version note: Students are encouraged to sketch functions whenever possible rather than always using graphical calculators. A translation has been used to obtain the function shown and the inequality is found using the sketch.
22 e.g. 2 Solve the inequality . The Modulus Functione.g. 2 Solve the inequalityMethod 1: Sketch the graphs andxyis above 1 in two regions.
23 e.g. 2 Solve the inequality . The Modulus Functione.g. 2 Solve the inequalityMethod 1: Sketch the graphs andxyis above 1 in two regions.The sketch doesn’t show the x-coordinates of A and B, so we must find them.ABxx13The right hand branch of the modulus graph is the same as y = x –so B is given byThe gradient of the right hand branch is +1 and the left hand is -1, so the graph is symmetrical.So,
24 18: Iteration Diagrams and Convergence Demo version note:By this stage the students understand how to rearrange equations in a variety of ways in order to find an iterative formula. They have met cobweb and staircase diagrams in simple cases of convergence and divergence.
25 Iteration Diagrams and Convergence In the next example we’ll look at an equation which has 2 roots and the iteration produces a surprising result.The equation isWe’ll try the simplest iterative formula first :If you have Autograph or another graph plotter you may like to try to find both roots before you see my solution.
26 Iteration Diagrams and Convergence Let’s try with close to the positive root.Let
27 The sequence diverges rapidly. Iteration Diagrams and ConvergenceLet’s try with close to the positive root.The staircase moves away from the root.The sequence diverges rapidly.Using the iterative formula,
28 Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.
29 Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.
30 Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.
31 Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.
32 Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.The sequence now converges but to the other root !
33 22: Integrating the Simple Functions Demo version note:The opportunity is taken here to remind the students about using integration to find areas, whilst applying the work to a trig integration.
34 Integrating the Simple Functions (b)Radians!The definite integral can give an area, so this result may seem surprising. However, the graph shows us why it is correct.This part gives a positive integralHow would you find the area?Ans: Find the integral from 0 toand double it.This part gives a negative integral
35 28: Integration giving Logs Demo version note:Inspection is used to find integrals of the formIn this example, the method is applied to a question where the integrand needs adjusting.
36 Integration giving Logs What do we want in the numerator?Ans:We now need to get rid of the minus and replace the 3.. . .
37 Integration giving Logs What do we want in the numerator?Ans:We now need to get rid of the minus and replace the 3.Always check by multiplying the numerator by the constant outside the integral:
38 Integration giving Logs What do we want in the numerator?Ans:We now need to get rid of the minus and replace the 3.
39 31: Double Angle Formulae Demo version note:Summaries are given at appropriate points in the presentations. The notebook icon suggests that students might want to copy the slide.
41 Double Angle FormulaeN.B. The formulae link any angle with double the angle.For example, they can be used forandandandandWe use themto solve equationsto prove other identitiesto integrate some functions
42 43: Partial Fractions Demo version note: Introductory exercises follow most sections of theory and examples. The next slide shows the first exercise on finding Partial Fractions. The full solution is given and the cover-up method used to check the result. Students would use their textbooks for further practice.
43 Partial FractionsExercisesExpress each of the following in partial fractions.184.108.40.206.
44 Partial FractionsSolutions:1.Multiply by :So,Check:
45 Partial FractionsSolutions:1.Multiply by :So,Check:
46 Partial FractionsSolutions:1.Multiply by :So,Check:( You don’t need to write out the check in full )
47 47: Solving Differential Equations Demo version note:The slides here show the introduction to the method of separating the variables to solve some differential equations. Solutions are applied later to applications of growth and decay functions.
48 Solving Differential Equations e.g. (2)Before we see how to solve the equation, it’s useful to get some idea of the solution.The equation tells us that the graph of y has a gradient that always equals y.We can sketch the graph by drawing agradient diagram.For example, at every point where y = 2, the gradient equals 2. We can draw a set of small lines showing this gradient.21We can cover the page with similar lines.
49 Solving Differential Equations We can now draw a curve through any point following the gradients.
50 Solving Differential Equations However, we haven’t got just one curve.
51 Solving Differential Equations The solution is a family of curves.Can you guess what sort of equation these curves represent ?ANS: They are exponential curves.
52 51: The Vector Equation of a Line Demo version note:By this time, the students have practised using vectors and are familiar with the notation. The equation of a straight line in vector form is developed.
53 The Vector Equation of a Line Finding the Equation of a LineIn coordinate geometry, the equation of a line ise.g.The equation gives the value (coordinate) of y for any point which lies on the line.The vector equation of a line must give us the position vector of any point on the line.We start with fixing a line in space.We can do this by fixing 2 points, A and B. There is only one line passing through these points.
54 A and B are fixed points. a r1 The Vector Equation of a Line We consider several more points on the line.xar1We need an equation for r, the position vector of any point R on the line.xStarting with R1:
55 A and B are fixed points. a r2 The Vector Equation of a Line We consider several more points on the line.xaWe need an equation for r, the position vector of any point R on the line.xxr2Starting with R1:
56 A and B are fixed points. r3 a The Vector Equation of a Line We consider several more points on the line.r3xaWe need an equation for r, the position vector of any point R on the line.xxStarting with R1:
57 t is called a parameter and can have any real value. The Vector Equation of a LinexSo for R1, R2 and R3xxaxxFor any position of R, we havet is called a parameter and can have any real value.It is a scalar not a vector.
58 Full version available from:- Chartwell-Yorke Ltd. 114 High Street, Belmont Village,Bolton,Lancashire,BL7 8ALEngland, tel (+44) (0) , fax (+44) (0)
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