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Mathematical Methods (CAS) Unit 3 VCE SOLUTIONS HEADSTART WORKSHOP Functions and relations Set Notation [ or ] ‘Closed brackets’ – including that end-point [2,3] means every real number between and including 2 and 3 ( or ) ‘Open brackets’ – excluding that end-point When describing a set with end-point ∞ or -∞, make sure to use the ‘open brackets’ as opposed to the closed brackets. (2,3) means every real number between but not including 2 or 3 {…} A set of numbers containing the specific terms as listed within the brackets {1, 5, 19} means the set of numbers 1, 5 and 19 \ R\[2,3] means every real number excluding the set of numbers between 2 and 3, and also excluding 2 and 3 themselves R\{2,3} means every real number excluding 2 and 3 ∩ Intersection of two sets – gives the set which contains all elements common to both of the two original sets [-5,7] ∩ (2,10) notates the intersection of the sets [-5,7] and (2,10) and is defined by the set (2,7] U Union of the two sets – gives the set which contains all elements contained in either of the two original sets [-5,7] U (2,10) notates the union of the sets [-5,7] and (2,10) and is defined by the set [-5,10) ∈ ‘is an element of the set described following’ 1 ∈ R means that 1 is an element of the set of all real numbers ⊆ ‘is a subset of the set described following’ – i.e. all the elements of the set preceding the symbol are contained in the set following the symbol (1,5) ⊆ (-1,7) means that all the elements of the set (1,5) are contained within the set (-1,7) Types of functions ‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒ Linear functions Quadratic functions Cubic functions Quartic functions Inverse functions Composite functions Hybrid Functions Modulus functions Discriminant and Determinant Sketching functions Linear and Quartic – very rarely examined ‒ Know how the shapes of each quartic graph ‒ In your book Quadratic and Cubic – sketching mostly relates to transformations and integration/differentiation ‒ Sketching quadratic and cubic functions are essential to know ‒ Know the shapes and how to find turning points ‒ Different ways to write the equations Sketching functions A few basic tips: ◦ When sketching any graph, a good place to start is always to find any x and y axis intercepts. ◦ For example, If you need to sketch a linear graph, all you need to do is find the x and y axis intercepts, and then draw a line through those points. ◦ As the degree (the degree is essentially the value of the highest power in a polynomial function – i.e. x^3 + x^2 +2 would be of degree 3, and x + 2 would be of degree 1) of a polynomial increases, polynomials generally require a few more key points to be found. ◦ For instance, a quadratic function (parabola) has an turning point that needs to be found. A cubic function may have two turning points that need to be found, and a quartic may have three. But in each case, the first steps can simply be finding the x and y ints. You will learn how to find turning points when you start calculus – it’s not too hard, but for now all you need to do is find x & y ints! ◦ But to find x and y ints polynomials, you’ll need to know how to factorise themwell Factorising polynomials What is the process of carrying out polynomial long division? In maths methods, long division is used to divide a polynomial by another polynomial, where the degree of the numerator is equal to or greater than the degree of the denominator. Sometimes dividing through will result with a remainder Example: (2x2-5x-1)/(x-3) And sometimes it will result with no remainder Example: (x2-3x-4)/(x-4) When there is no remainder, it can be easier to use another method (instead of long division) to find the unknown factor: Example: (x2-3x-4)/(x-4) What is the remainder theorem? What is the factor theorem? if x-a if a factor, then P(a) = r (where r is the remainder) What is the factor theorem? if x-a if a factor, then P(a) = 0 - Whether there is a remainder or no remainder depends on the polynomial. Sketching Polynomials Now that we know how to factorise polynomials, we can go about sketching them. Just as you learnt in year 10 and 11 that it is easier to sketch a parabola when it is in the form (x-a)(x-b) as compared to ax^2 + bx + c, it’s the same for polynomials of higher degree This is because we can let y=0 to find the x-intercepts, and by the null factor law, the x-ints are essentially the negative components of each factor; that is, for f(x) = (x-a)(x-b)(x-c), the x-ints are x=a, x=b, x=c (as we put f(x) = 0). And once we’ve found these x-ints, finding the y-ints is always a lot easier Therefore, being able to sketch polynomials is more about being able to factorise them than anything else. Of course, there are a few other considerations, such as whether it is positive or negative, or translated, etc (another key consideration is the turning points, but you learn this later on in calculus). But if you know how to factorise polynomials well, then you can basically sketch their graph. Note: this is why the quick method for factorising polynomials is so useful – to sketch graphs faster. Square root functions Firstly, whenever you see a square root function, you must be very conscious that you can’t take the square root of a negative number. This is why the standard sq root functions only exist for positive x-values (and zero). Knowing sq root functions only exist when the “inside” part is greater than zero, we can easily establish our implied domain for the function Example: f(x) = As always, it’s a good idea to find x and y intercepts Other things to know: if there’s a negative sign in front of the sq root, then the graph will be flipped upside down, and if there’s a negative sign in front of the ‘x’ term inside the sq root, then the graph will be flipped side ways. You will understand this better once you go through transformations. Then you can sketch. However, it also helps to find one other point to help guide your sketch (i.e. you could substitute x=1 into f(x) to find the y-value, and then you’d have an extra point to play around with) Hyperbolae With sq root functions, we established that you must always remember that you can’t take the square root of a negative number. With hyperbolae, the condition to remember is that you can’t divide by zero. i.e. the denominator cannot equal zero Example: f(x) = This gives us the vertical asymptote. However, we also know that no matter how large x becomes, the overall quotient term will not equal zero. So we get a horizontal asymptote as well; that is, y=0 is the horizontal asymptote (if the hyberbola was translated +c units, the asymptote would y = c) Trunci Very similar to Hyperbolae – denominator cannot equal zero, so the process is the same. Since the denominator is squared, all negative components on the left become positive, just like on the right. Hence, the graph is symmetrical about its vertical asymptote. Matrices An (m x n) matrix multiplied by an (n x r) is defined as the number of columns in the first matrix equals the number of rows in the second matrix. The dimension of the resulting matrix is (m x r) That is (m x n) x (n x r) produces a (m x r) matrix. Inverse matrices: Firstly, the determinant of a matrix is det(A) = ad-bc for a matrix A= -You can think of the determinant as the “magnitude” of a matrix, but this magnitude will often turn out to be completely different to what you would expect -We need the determinant to be able to find the inverse of a matrix A, which is denoted A^-1: Using inverse matrices The most common matrix equation is Ax=B where A has dimensions (2x2) and B has dimensions (2x1). If we want to solve for x, we simply pre-multiply both sides of the equation by A^-1 (which is why the inverse matrix is useful) -(explain further on board) Once you understand the inverse matrix, you can apply it to more difficult problems. The most commonly examined application of the inverse matrix resonates with the following example. For this example, it is important to realise that solutions only exist when the inverse matrix exists, and the inverse on exists when the determinant isn’t equal to zero (as the determinant is on the denominator in the expression for the inverse – and you can’t divide by zero) Find the values of m for which there is a unique solution, and then infinitely many or no solutions. Inverse functions Only exists when function is one-to-one, that is it fulfils the requirements of both the horizontal and vertical line test Sketching these graphs – reflect about the line = To find the inverse functions must state: for inverse functions, switch and Common exam question asks you to find the maximal domain of () so that an inverse function exists Be aware that an inverse function and its original function ALWAYS intersect at = Example – Inverse functions Consider the function : (, 3) → , () = + 1 a) Find the smallest value for b) Find the inverse function 2 Composite functions ∘ () only exists when the range of is a subset of or equal to the domain of (i.e. ran ⊆ dom ) Common exam questions include finding the maximal domain of for a composite function to exist Good technique is to draw the table to find the suitable range of then by extension domain of g that allows ∘ () to exist Example – Composite Functions Let : → , = 4 − 2 and be the set of all real values of for which () is deﬁned. Let : → , where () = 2 + 1. Given ∘ () exists: a) Find the maximum domain for (), . b) Is ∘ () defined for the () with domain ? State reasons c) Sketch ∘ () Hybrid functions Simply, more than one function sketched on one set of axis ‒ Can be tied with kinematics and integration ‒ Cannot be differentiated when function changes (or at the cusp) Common questions asks you to sketch, find maximal domain and range (examined most commonly in the first exam) Sketch each part of the hybrid function as separate functions Modulus function is a special type of hybrid function Modulus functions Modulus is essentially the magnitude – hence all negatives become positive A type of composite function where one of the functions is () = || First type – takes the modulus of the entire function – ∘ Second type – takes the modulus of only the x value – ∘ () Common questions asks you to sketch, find maximal domains and ranges Example – Modulus functions Let () = 2 − 6 + 5 and () = || a) Find ∘ and sketch the graph and state the maximal domain and range. b) Find ∘ () and sketch the graph and state the maximal domain and range. Discriminant and determinant The discriminant is given by the expression Δ = 2 − 4 ‒ When it equals 0, there is one solution ‒ When it is greater than 0, there are multiple solutions ‒ When it is less than 0, there are no solutions The determinant of some matrix is given by the expression det = = − (from matrix) ‒ det() = 0, no solutions or infinitely many solutions exist ‒ det() ≠ 0, a unique solution exists Can use either to solve simultaneous equations Example Discriminant and determinant Consider the simultaneous equations + 2 = 8 and 4 − (2 − ) = 2 Find the value(s) of for which there are a) no solutions b) infinitely many solutions c) a unique solution Index laws × = + = − = × = 1 = = 1 − = 0 = 1 1 = Log laws log = log () + log () log = log () − log () log = log () log log () = log − log = log log 1 = 0 log = 1 1 Transformations Transformations =× + + ‒ is a dilation of factor from the axis (changing the value) ‒ is a dilation of factor from the axis (changing the value) ‒ is a translation of units in the negative direction of the axis ‒ is a translation of units in the positive direction of the axis ‒ If is negative, it is also a reflection in the axis ‒ If is negative, it is also a reflection in the axis Transformations Situation 1: Given a set of transformations to apply to a function, you are asked to find the new transformed function. 1 Example: = 2 is transformed by: ‒ a reflection in the axis ‒ a translation of 3 units in the positive direction of the axis ‒ a dilation of 2 units from the axis Write down the equation of the new transformed function. Transformations Situation 2: Given the transformed function and original function, you are asked to find the transformations performed to change the original function to the transformed function. Example: 1 = is transformed to () = 2 −1 +3 State the sequence of transformations required. Transformations Situation 3: Given a “complex” transformed function and you are asked to find the transformations you have to perform to map it back to the original “simple” function. Example: () = 3 – 7 is 1− 2 transformed to () = 1 2 State the sequence of transformations required Transformations using the dash method What are the steps to using the dash method? Transformation using matrices Remember top left box affects the axis and so is therefore a reflection/dilation in or from the axis Bottom right box affects the axis and so is therefore a reflection/dilation in or from the axis Transformations of the top right and bottom left boxes are rarely examined in this course but is a reflection in the line = . Translations are graphed on the 2 × 1 matrix. Thanks for coming to VCE Solutions’ first workshop of the year! Feel free to send in any VCE-related questions you have over Facebook – just message our page. Summa Wu will be running weekly classes on Monday afternoons – 4:306pm here (Hawthorn Library). You can register for our classes at www.vcesolutions.com.au, and they’re only $30 for an hour and a half – full notes and slides will also be given each week. We will also have weekly Biology classes held by Winston Dzau on Sunday afternoons (2-3:30pm) for the same price. Our study guides will be available on our website, and are written to help you achieve an awesome score in your subject – these are the actual notes used by our authors when studying VCE helping them achieve perfect scores. Study Guides available at the moment are Methods, Psychology and Biology which are $35 each. We hope you enjoyed today’s session, and thanks for coming!