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Engineering Analysis 1 : Elementary Functions

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Engineering Analysis 1 : Elementary Functions Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Engineering Analysis 1 : Elementary Functions Dr. Paul D. Ledger, Dr Rob Daniels, Dr Igor Sazonov [email protected] College of Engineering, Swansea University, UK PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Outline 1 Functions 2 Polynomial Functions 3 Rational Functions 4 Circular Functions 5 Exponential and Logarithmic Functions 6 Continuous and Discontinuous Functions PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Function Definitions A function expresses how the dependent variable depends on the independent variable. In engineering both variables involved usually have some physical meaning. A function consists of two different sets of numbers X and Y and rule that assigns to each element x in the set X precisely one element y from the set Y. We denotes this by f : x ! y or y = f (x). We say that x is the argument of f . The set X is called the domain of the function. The set Y is called the codomain of the function. Rule X Y Note that there may be elements of Y that are not outputs of the function. The set of all images y = f (x), x in X is called the image set or range of f Hence the range is not always the same as the codomain. PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Function Properties It is worth noting the following points In the definition of the function, each input gives rise to exactly one output. A function which has the special property that different inputs gives rise to different outputs is said to be one to one or injective. A function which has that has the special property that every element of its codomain is an output (ie whose range is the whole of its codomain) is said to be on to or surjective. A function which is both injective and surjective is called bijective. It is important to understand a functions properties in order to be able construct its inverse. PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Function Examples 2 x+3 x2 40 15 35 10 30 25 5 y y 20 0 15 10 -5 5 -10 0 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 x x y = f (x) = 2x + 3 y = f (x) = x2 Domain is R Domain is R Codomain is R Codomain is R + Range is R Range is R = fy : y 2 R; y ≥ 0g Function is injective and surjective ie Function is neither injective or is bijective surjective. ie not bijective. PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Inverse Functions For certain y = f (x) it is possible to determine what values of x give rise to a value of y (ie construct its inverse). In general if a function y = f (x) is bijective (ie injective and surjective)the inverse function x = f −1(y) can be constructed (often by simple rearrangement). Sometimes, if it does not lead to ambiguity, we can interchange x and y and write the inverse function as f −1(x) If a function is not bijective, we may need to change the domain and/or codomain to make it bijective. Example −1 y−b For the bijective function y = f (x) = ax + b with a 6= 0 its inverse is x = f (y) = a −1 x−b and we could write f (x) = a PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Function Examples y = f (x) = x2 x2 x2 100 100 90 90 80 80 70 70 60 60 y 50 y 50 40 40 30 30 20 20 10 10 0 0 -10 -8 -6 -4 -2 0 0 2 4 6 8 10 x x − + Domain is R Domain is R + + Codomain is R Codomain is R + + Range is R Range is R Function is injective and surjective ie Function is injective and surjective ie is bijective is bijective p p x = f −1(y) = − y x = f −1(y) = y PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Outline 1 Functions 2 Polynomial Functions 3 Rational Functions 4 Circular Functions 5 Exponential and Logarithmic Functions 6 Continuous and Discontinuous Functions PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Polynomials A polynomial function has the general form n n−1 f (x) = anx + an−1x + ··· + a1x + a0 r where n is the a positive integer, ar is a real number called the coefficient of x , r = 0; 1; ··· ; n. The index n of the highest power of x occurring in f (x) is called the degree. For n = 1 we obtain the linear function f (x) = a1x + a0 for n = 2 we get the quadratic function 2 f (x) = a2x + a1x + a0 PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Graphs of polynomial functions 100 2x-3 x2 -3x+4 x2-10 30 x2-5*x+20 -x2/2+2*x 20 50 10 y 0 0 -10 -20 -30 -50 -10 -5 0 5 10 -10 -5 0 5 10 x x 2 y = f (x) = a1x + a0 y = f (x) = a2x + a1x + a0 a0 is the intercept on y axis a0 is the intercept on y axis ∆f ∆y a controls the vertex. By completing a1 = ∆x = ∆x is the gradient 1 the square the vertex is a2− a a − a1 ; − 1 4 0 2 . 2a2 4a2 a2 controls whether the quadratics opens upward (a2 > 0) or downwards (a2 < 0) and also the speed of increase of the quadratic PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Properties of polynomial functions If two polynomials are equal for all values of the independent variable, then coefficients of the powers of the variable are equal. Thus if n n−1 f (x) = anx + an−1x + ··· + a1x + a0 n n−1 g(x) = bnx + bn−1x + ··· + b1x + b0 and f (x) = g(x) for all x then ai = bi for i = 0; 1; ··· ; n. Any polynomial with real coefficients can be expressed as a product of linear and irreducible quadratic factors (an irreducible quadratic factor is one that cannot be factored into the product of two linear terms with real coefficients). PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Example Question Find values of A, B and C to ensure that x2 + 1 = A(x − 1) + B(x + 2) + C(x2 + 2) for all values of x. Solution We first gather the appropriate terms together on the right hand side 1x2 + 1 = Cx2 +( A + B)x +( −A + 2B + 2C) By comparing and equating coefficients of x2, x1 and x0 we find C = 1 A + B = 0 − A + 2B + 2C = 1 which lead to the result 1 1 A = B = − C = 1 3 3 PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Example Question Find values of A, B and C to ensure that x2 + 1 = A(x − 1) + B(x + 2) + C(x2 + 2) for all values of x. Solution We first gather the appropriate terms together on the right hand side 1x2 + 1 = Cx2 +( A + B)x +( −A + 2B + 2C) By comparing and equating coefficients of x2, x1 and x0 we find C = 1 A + B = 0 − A + 2B + 2C = 1 which lead to the result 1 1 A = B = − C = 1 3 3 PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Example Question factorise the polynomial x3 − 3x2 + 6x − 4 = 0 Solution The function f (x) = x3 − 3x2 + 6x − 4 is equal to zero when x = 1. We say that x = 1 is a zero of the function. Thus x − 1 must be a factor of f (x). We factor out x − 1 as follows x3 − 3x2 + 6x − 4 =( x − 1)(x2 − 2x + 4) To see if we can simplify this further, we need to find if g(x) = x2 − 2x + 4 has linear factors with real coefficients. To check this we check whether g(x) = 0 has real roots using 2 ± p4 − 4(4) x = 2 p p In this case we see that are no real roots as 4 − 16 = −12 is not defined. Therefore (x2 − 2x + 4) is an irreducible quadratic term and our factorised solution is x3 − 3x2 + 6x − 4 = (x − 1)(x2 − 2x + 4) PDL, RD, IS (CoE) WS 2016 Functions Polynomial Functions Rational Functions Circular Functions Exponential and Logarithmic Functions Continuous and Discontinuous Functions Example Question factorise the polynomial x3 − 3x2 + 6x − 4 = 0 Solution The function f (x) = x3 − 3x2 + 6x − 4 is equal to zero when x = 1. We say that x = 1 is a zero of the function.
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