MATHS WORKSHOPS. Functions. Business School
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1 MATHS WORKSHOPS Functions Business School
2 Outline Overview of Functions Quadratic Functions Exponential and Logarithmic Functions Summary and Conclusion
3 Outline Overview of Functions Quadratic Functions Exponential and Logarithmic Functions Summary and Conclusion
4 Functions Definition (Function) A function, f, is a mapping from one value, X, to another value, Z: f : X Z. Think of the function, f, as a machine that takes an input, X, then transforms it in some way and outputs the result: Z. More Example (f(x) = 2x 3) The function f(x) = 2x 3 has the following mapping: f(3) = = 3 so f(x) maps 3 to 3. f(2) = = 1 so f(x) maps 2 to 1. f(1) = = 1 so f(x) maps 1 to 1. f(0) = = 3 so f(x) maps 0 to 3.
5 Functions Definition (Function) A function, f, is a mapping from one value, X, to another value, Z: f : X Z. Think of the function, f, as a machine that takes an input, X, then transforms it in some way and outputs the result: Z. More Example (Where have we seen functions before?) We ve been working with functions already: Linear Functions: f(x) = ax + b Quadratic Functions: f(x) = ax 2 + bx + c (today s lesson) Sometimes functions are written as y = f(x). For example you may see, y = ax + b instead of f(x) = ax + b.
6 Outline Overview of Functions Quadratic Functions Exponential and Logarithmic Functions Summary and Conclusion
7 Quadratic Function Definition (Quadratic Function) A quadratic function takes the form: f(x) = ax 2 + bx + c where a, b and c are parameters and x is a variable. More Key point The key point is that there is a term in the function involving the square of the variable, x 2. Definition (Parabola) The graph of a quadratic functions is a curve often referred to as a parabola. More
8 Factorising quadratic function Definition (Factorisation) Quadratic functions can be factorised to take the form: f(x) = (x + d)(x + e) = (x + d)x + (x + d)e = x 2 + dx + xe + de (factorised) = x 2 + (d + e)x + de (expanded) More Example (Expand the following factorised quadratic) (x 2)(x + 1) =
9 Factorising quadratic function Definition (Factorisation) Quadratic functions can be factorised to take the form: f(x) = (x + d)(x + e) = (x + d)x + (x + d)e = x 2 + dx + xe + de (factorised) = x 2 + (d + e)x + de (expanded) More Example (Expand the following factorised quadratic) (x 2)(x + 1) = (x 2)x + (x 2) 1
10 Factorising quadratic function Definition (Factorisation) Quadratic functions can be factorised to take the form: f(x) = (x + d)(x + e) = (x + d)x + (x + d)e = x 2 + dx + xe + de (factorised) = x 2 + (d + e)x + de (expanded) More Example (Expand the following factorised quadratic) (x 2)(x + 1) = (x 2)x + (x 2) 1 = x 2 2x + x 2 = x 2 x 2
11 Factorising quadratic functions Example (Factorise f(x) = x x + 32) We factorise this to (x + d)(x + e) by matching the expansion: x 2 + (d + e)x + de with our example: x x I.e. we try to find factors d and e such that: P Their product is de = 32 so that the constants match. S Their sum is d + e = 12 so the coefficients of x match. F The factors are the two numbers whose sum is 12 and their product is 32. By trial and error we notice that = 12 and 4 8 = 32: f(x) = (x + 4)(x + 8).
12 Factorising a more complex quadratic The general form for a quadratic function is: f(x) = ax 2 + bx + c Example (Factorise f(x) = 2x 2 + 3x 5) Here we want to break up the middle term to help factorisation. To do this we find two numbers whose: P product is ac = 2 5 = 10 S sum is b = 3 F Using trial and error we find suitable candidates 2 and 5. We can then re-write the original equation as: 2x 2 + 3x 5 = 2x 2 2x + 5x 5 = 2x(x 1) + 5(x 1) = (2x + 5)(x 1)
13 Your turn to factorise... Factorise this expression x 2 + 5x + 6 in the form (x + d)(x + e). P Product: S Sum: F Factors:
14 Your turn to factorise... Factorise this expression x 2 + 5x + 6 in the form (x + d)(x + e). P Product: Need two numbers that multiply to give 6 S Sum: Need two numbers that sum to give 5 F Factors:
15 Your turn to factorise... Factorise this expression x 2 + 5x + 6 in the form (x + d)(x + e). P Product: Need two numbers that multiply to give 6 S Sum: Need two numbers that sum to give 5 F Factors: = 5 and 2 3 = 6
16 Your turn to factorise... Factorise this expression x 2 + 5x + 6 in the form (x + d)(x + e). P Product: Need two numbers that multiply to give 6 S Sum: Need two numbers that sum to give 5 F Factors: = 5 and 2 3 = 6 Therefore the factorisation is: (x + 2)(x + 3) You can verify this by expanding it out again!
17 Factorising using the cross method In general, we can factorised ex 2 + fx + g using acx 2 + (ad + bc)x + bd = (ax + b)(cx + d) by finding a, b, c, d such that ac = e, bd = g and ad + bc = f, using the cross method: ax b check if ad + cb = f cx d 1. Pick an a and a c such that ac = e 2. Pick a b and a d such that bd = g 3. If ad + cb = f then you have the solution. If ad + cb f go back to Step 1.
18 Your turn to factorise with a harder example... 2x x + 5 Method 1: Using a PSF-type approach: P Product: S Sum: F Factors:
19 Your turn to factorise with a harder example... 2x x + 5 Method 1: Using a PSF-type approach: P Product: Need two numbers that multiply to give 2 5 = 10 S Sum: Need two numbers that sum to give 11 F Factors:
20 Your turn to factorise with a harder example... 2x x + 5 Method 1: Using a PSF-type approach: P Product: Need two numbers that multiply to give 2 5 = 10 S Sum: Need two numbers that sum to give 11 F Factors: = 11 and 10 1 = 10
21 Your turn to factorise with a harder example... 2x x + 5 Method 1: Using a PSF-type approach: P Product: Need two numbers that multiply to give 2 5 = 10 S Sum: Need two numbers that sum to give 11 F Factors: = 11 and 10 1 = 10 We can use this to re-write the original expression: 2x x + 5 = 2x 2 + x + 10x + 5 = x(2x + 1) + 5(2x + 1) = (x + 5)(2x + 1).
22 Your turn to factorise with a harder example... 2x x + 5 Method 2: the cross method to factorise this as: acx 2 + (ad + bc)x + bd = (ax + b)(cx + d) Here we want to find a, b, c, d such that ac = 2, bd = 5 and ad + bc = 11. ax = x b = 5 cx = 2x d = 1 1. Pick a = 1 and c = 2 such that ac = 2 2. Pick b = 5 and d = 1 such that bd = 5 3. Check if ad + cb = 11. Here = 11, so we have a solution: (x + 5)(2x + 1)
23 Graphing quadratic functions One way to graph quadratic functions would be to plot some points and join them. Consider the function, f(x) = x 2 : x f(x) = x 2 2 ( 2) 2 = ( 1.5) 2 = 2.25 f(x) 1 ( 1) 2 = ( 0.5) 2 = = (0.5) 2 = = (1.5) 2 = = x 4
24 Graphing quadratic functions One way to graph quadratic functions would be to plot some points and join them. Consider the function, f(x) = x 2 : x f(x) = x 2 2 ( 2) 2 = ( 1.5) 2 = 2.25 f(x) 1 ( 1) 2 = ( 0.5) 2 = = (0.5) 2 = = (1.5) 2 = = x 4 x 2
25 Graphing quadratic functions Consider quadratic functions of the form y = ax 2. What does a do? f(x) x 2 x
26 Graphing quadratic functions Consider quadratic functions of the form y = ax 2. What does a do? f(x) 3x 2 x 2 x
27 Graphing quadratic functions Consider quadratic functions of the form y = ax 2. What does a do? f(x) 3x 2 x x2 x a changes the steepness of the curve.
28 Graphing quadratic functions Consider quadratic functions of the form y = ax 2. f(x) x 1 2 x2 3x 2 x 2 The sign of a determines whether the parabola is convex (smile) or concave (frown). More
29 Graphing quadratic functions Consider quadratic functions of the form f(x) = x 2 + c. What does c do? f(x) x x 0.5
30 Graphing quadratic functions Consider quadratic functions of the form f(x) = x 2 + c. What does c do? f(x) x x x 0.5
31 Graphing quadratic functions Consider quadratic functions of the form f(x) = x 2 + c. What does c do? f(x) 1.5 x x 2 x x 0.5 c moves the curve up and down.
32 Graphing quadratic functions Consider quadratic functions of the form f(x) = ax 2 + bx + c What does b do? f(x) x 2 + 3x x -1-2
33 Graphing quadratic functions Consider quadratic functions of the form f(x) = ax 2 + bx + c What does b do? f(x) x 2 + 3x 2 1 x 2 3x x -1-2
34 Graphing quadratic functions Consider quadratic functions of the form f(x) = ax 2 + bx + c What does b do? f(x) 2 1 x 2 + 3x x 2 3x x 2 3x x -1-2 b moves the curve from side to side
35 Finding the roots graphically Definition (Roots) The point(s) at which the quadratic function crosses the x axis are called the roots of the function. More Example (Finding the roots of f(x) = x 2 + 3x graphically) -3-2 f(x) x 2 + 3x x
36 Finding the roots graphically Definition (Roots) The point(s) at which the quadratic function crosses the x axis are called the roots of the function. More Example (Finding the roots of f(x) = x 2 + 3x graphically) -3-2 f(x) x 2 + 3x x The roots occur at x = 3 and x = 0.
37 Finding the roots algebraically The definition says that the roots are The point(s) at which the quadratic function crosses the x axis. Mathematically this is when f(x) = 0. This is easiest to find using the factorised form of f(x). Example (Finding the roots of f(x) = x 2 + 3x) 1. Factorise f(x): f(x) = x 2 + 3x = x(x + 3). 2. Work out the values of x for which f(x) = 0 is true. When x = 0 then x(x + 3) = 0. When x = 3 then x(x + 3) = 0. Therefore the roots are x = 0 and x = 3.
38 What if there aren t any roots? The function x 2 + 2x + 1 doesn t cross the x axis at all! f(x) x 2 + 2x x Definition (Discriminant) The function f(x) = ax 2 + bx + c will only have root(s) if b 2 4ac 0. b 2 4ac is known as the discriminant. More
39 The Quadratic Formula Definition (Quadratic Formula) If the quadratic function f(x) = ax 2 + bx + c has roots, they can always be found using the quadratic formula: x = b ± b 2 4ac. More 2a Note that the square root of the discriminant is in the quadratic formula. This result suggests why there are no real roots unless b 2 4ac 0. You cannot take the square root of a negative number. 1 1 You actually can but the solution is an imaginary number! More
40 The Quadratic Formula Example (Finding the roots of f(x) = x 2 + 3x) Here a = 1, b = 3 and c = 0 so x = b ± b 2 4ac 2a = 3 ± = 3 ± = 3 ± 3 2 = AND = 0 AND f(x) x 2 + 3x 1 2 x
41 Outline Overview of Functions Quadratic Functions Exponential and Logarithmic Functions Summary and Conclusion
42 Other common functions Example (Exponential function) f(x) = ae cx+b f(x) e 2x e x 1 2 ex where a, b, c and e are constants 1 x The graph of f(x) = e x is upward-sloping, and increases faster as x increases. The graph is always above the x-axis but gets arbitrarily close to it for negative x: the x-axis is an asymptote. More More
43 Other common functions Example (Logarithmic function) f(x) = a log(cx+b) where a, b and c are constants f(x) 1 log(2x) log(x) 1 2 log(x) x The graph of f(x) = log(x) slowly grows to positive infinity as x increases. More The graph is always to the right of the y-axis but gets arbitrarily close to it for small x: the y-axis is an asymptote.
44 Relationship between logs and exponentials In y = log a (b), a is known as the base of the log. We can change the base of the log using the relationship: log a (b) = log c(b) log c (a). Using this relationship, it is clear that: If y = log a (x) then x = a y log a (a) = log c(a) log c (a) = 1. Equivalently, if y = a x then x = log a y. If the base is the same, then the log function is the inverse of the exponential function: a log a (x) = x just like ax a = x.
45 Natural Logs We typically work with log base e log e (x) is often written as ln(x). If the base is left off the log, it s assumed log(x) = log e (x) (unless it s on your calculator, in which case it means log 10 ) Note that the usual relations hold: y = e x ln(y) = ln(e x ) ln(y) = x (taking log e of both sides) Also, y = ln(x) e y = e ln(x) e y = x (exponentiating both sides)
46 Exponential and Log rules Exponentiation: a 0 = 1 for all a 0. a 1 = 1 a a x a y = a x+y Logarithms: log(x y ) = y log(x) ax a y = ax y (a x ) y = a xy (ab) x = a x b x log(xy) = log(x) + log(y) ( ) x log = log(xy 1 ) = log(x) + log(y 1 ) = log(x) log(y) y log(1) = 0 More More
47 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = 2. log 2 y 3 = 4 = y = Simplify the following expressions 1. e ln 5 = 2. ln e = 3. e x+ln x = 4. ln(1 + x) ln(1 x) = ln(1 + x) 5. ln(e 2 = ) 6. log 3 3 q =
48 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = Simplify the following expressions 1. e ln 5 = 2. ln e = 3. e x+ln x = 4. ln(1 + x) ln(1 x) = ln(1 + x) 5. ln(e 2 = ) 6. log 3 3 q =
49 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = 3 24 = 48 Simplify the following expressions 1. e ln 5 = 2. ln e = 3. e x+ln x = 4. ln(1 + x) ln(1 x) = ln(1 + x) 5. ln(e 2 = ) 6. log 3 3 q =
50 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = 3 24 = 48 Simplify the following expressions 1. e ln 5 = 5 2. ln e = 3. e x+ln x = 4. ln(1 + x) ln(1 x) = ln(1 + x) 5. ln(e 2 = ) 6. log 3 3 q =
51 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = 3 24 = 48 Simplify the following expressions 1. e ln 5 = 5 2. ln e = ln e 1/2 = 1 2 ln e = e x+ln x = 4. ln(1 + x) ln(1 x) = ln(1 + x) 5. ln(e 2 = ) 6. log 3 3 q =
52 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = 3 24 = 48 Simplify the following expressions 1. e ln 5 = 5 2. ln e = ln e 1/2 = 1 2 ln e = e x+ln x = e x e ln x = e x x 4. ln(1 + x) ln(1 x) = ln(1 + x) 5. ln(e 2 = ) 6. log 3 3 q =
53 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = 3 24 = 48 Simplify the following expressions 1. e ln 5 = 5 2. ln e = ln e 1/2 = 1 2 ln e = e x+ln x = e x e ln x = e x x ( ) 1 + x 4. ln(1 + x) ln(1 x) = ln 1 x ln(1 + x) 5. ln(e 2 = ) 6. log 3 3 q =
54 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = 3 24 = 48 Simplify the following expressions 1. e ln 5 = 5 2. ln e = ln e 1/2 = 1 2 ln e = e x+ln x = e x e ln x = e x x ( ) 1 + x 4. ln(1 + x) ln(1 x) = ln 1 x ln(1 + x) ln(1 + x) 5. ln(e 2 = = 1 ln(1 + x) ) 2 ln(e) 2 6. log 3 3 q =
55 Your Turn... Solve the following equations for x 1. ln(x) = 2 = x = e 2 2. log 2 y 3 = 4 = y = 3 24 = 48 Simplify the following expressions 1. e ln 5 = 5 2. ln e = ln e 1/2 = 1 2 ln e = e x+ln x = e x e ln x = e x x ( ) 1 + x 4. ln(1 + x) ln(1 x) = ln 1 x ln(1 + x) ln(1 + x) 5. ln(e 2 = = 1 ln(1 + x) ) 2 ln(e) 2 6. log 3 3 q = q log 3 3 = q
56 Solve the following equations for x 1. ln(2x + 1) = ln(10 x) 3. ln(2x + 3) = x+1 = 4 x 4. 5 x+1 = 200
57 Solve the following equations for x 1. ln(2x + 1) = ln(10 x) exp{ln(2x + 1)} = exp{ln(10 x)} 2x + 1 = 10 x x = x+1 = 4 x 3. ln(2x + 3) = x+1 = 200
58 Solve the following equations for x 1. ln(2x + 1) = ln(10 x) exp{ln(2x + 1)} = exp{ln(10 x)} 2x + 1 = 10 x x = x+1 = 4 x 2 3x+1 = 2 2x log 2 (2 3x+1 ) = log 2 (2 2x ) 3x + 1 = 2x x = 1 3. ln(2x + 3) = x+1 = 200
59 Solve the following equations for x 1. ln(2x + 1) = ln(10 x) exp{ln(2x + 1)} = exp{ln(10 x)} 2x + 1 = 10 x x = x+1 = 4 x 2 3x+1 = 2 2x log 2 (2 3x+1 ) = log 2 (2 2x ) 3x + 1 = 2x x = 1 3. ln(2x + 3) = 3 exp{ln(2x + 3)} = e 3 2x + 3 = e x+1 = 200 x = e3 3 2
60 Solve the following equations for x 1. ln(2x + 1) = ln(10 x) exp{ln(2x + 1)} = exp{ln(10 x)} 2x + 1 = 10 x x = x+1 = 4 x 2 3x+1 = 2 2x log 2 (2 3x+1 ) = log 2 (2 2x ) 3x + 1 = 2x x = 1 3. ln(2x + 3) = 3 exp{ln(2x + 3)} = e 3 2x + 3 = e x+1 = 200 x = e3 3 2 ln(5 x+1 ) = ln(200) (x + 1) ln(5) = ln(200) x + 1 = ln(200) ln(5) x = ln(200) 1 ln(5)
61 Outline Overview of Functions Quadratic Functions Exponential and Logarithmic Functions Summary and Conclusion
62 Summary parameters and variables substitution and solving equations dependent vs independent variable linear function: f(x) = y = ax + b slope (is it a parameter or a variable?) intercept (is it a parameter or a variable?) finding the equation for a linear function given a plot plotting a linear function given an equation recognising quadratic functions factorising and expanding quadratics finding the roots of a quadratic using x = b ± b 2 4ac 2a graphs of quadratic equations (parabolas)
63 Coming up... Week 5: Simultaneous Equations and Inequalities Algebraic and graphical solutions to simultaneous equations Understanding and solving inequalities Week 6: Differentiation Theory and rules of Differentiation Differentiating various functions and application of Differentiation
64 Additional Resources Test your knowledge at the University of Sydney Business School MathQuiz: Additional resources on the Maths in Business website sydney.edu.au/business/learning/students/maths The University of Sydney Mathematics Learning Centre has a number of additional resources: Basic concepts in probability notes More Sigma notation notes More Permutations and combinations notes More There s also tonnes of theory, worked questions and additional practice questions online. All you need to do is Google the topic you need more practice with! More
65 Acknowledgements Presenters and content contributors: Garth Tarr, Edward Deng, Donna Zhou, Justin Wang, Fayzan Bahktiar, Priyanka Goonetilleke. Mathematics Workshops Project Manager Jessica Morr from the Learning and Teaching in Business. Valuable comments, feedback and support from Erick Li and Michele Scoufis. Questions, comments, feedback? Let us know at
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