SIDE NOTES: Symmetry, Transformations and Compositions Math 157 Calculus I for the Social Sciences Lecture 3 based on “Calculus Early Transcendentals Differential & Multi-Variable Calculus for Social Sciences” Symmetry, Transformations and Compositions This text, including the art and illustrations, are available under the Creative Commons license (CC BY-NC-SA), allowing anyone to reuse, revise, remix and redistribute the text. For more info visit http://creativecommons.org/licenses/bync-sa/4.0/ Calculus Early Transcendentals Differential & Multi-Variable Calculus for Social Sciences was adapted by Petra Menz and Nicola Mulberry from Lyryx’ lecture notes by Michael Cavers licensed under a Creative Commons Attribution Non-Commercial Share-Alike 3.0 Unported License (CC BY-NC-SA 3.0). The adaptation ensures congruency between the text Calculus Early Transcendentals Differential & Multi-Variable Calculus for Social Sciences and the lectures for MATH 157. 1 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Math 157 Calculus I for the Social Sciences Lecture 3 Symmetry, Transformations and Compositions Learning Outcomes – Detailed: The successful student will be able to 1. Transform elementary functions. 2. Model supply, demand, cost, revenue and profit using elementary functions. 3. Perform algebra with functions. 4. Determine the equilibrium point when working with supply and demand. 5. Determine the break-even point when working with cost, revenue and profit. 2 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Even and Odd Functions A function f is an even function if f (−x) = f (x), for all x in the domain of f . If you reflect the graph about the y -axis, the graph stays the same! A function f is an odd function if f (−x) = −f (x), for all x in the domain of f . If you rotate the graph 180◦ about the origin, the graph stays the same! y y y = cos(x) −2π − 3π 2 −π − π2 π 2 π x 3π 2 y = sin(x) 1 1 −2π − 3π 2 −π − π2 π 2 π x 3π 2 −1 −1 function function 3 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Transformations One way to create a new function from a known function is to apply a sequence of transformations. Types of transformations include: I I I Translations Stretches Reflections Function F (x) = f (x) + c F (x) = f (x) − c F (x) = f (x + c) F (x) = f (x − c) F (x) = af (x) F (x) = f (bx) F (x) = −f (x) F (x) = f (−x) F (x) = |f (x)| Condition c>0 c>0 c>0 c>0 a>0 b>0 How to sketch F (x) given the graph of f (x) Shift f (x) upwards by c units Shift f (x) downwards by c units Shift f (x) to the left by c units Shift f (x) to the right by c units Stretch f (x) vertically by a factor of a Stretch f (x) horizontally by a factor of 1/b Reflect f (x) about the x-axis Reflect f (x) about the y -axis Take the part of the graph of f (x) that lies below the x-axis and reflect it about the x-axis 4 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Performing Transformations Example √ Use appropriate transformations to sketch the graph of the function y = | x + 2 − 1| − 1. Solution: Basic graph: 2 y Horizontal translation: 2 Vertical translation: y 2 y 1 1 1 x −3 −2 −1 1 2 −3 −2 −1 −1 −2 Absolute value: x x 3 2 1 2 −3 3 −2 −1 −1 −1 −2 −2 Vertical translation: 2 −1 3 y 1 x −2 2 y 1 −3 1 1 2 3 x −3 −2 −1 1 −1 −1 −2 −2 2 3 5 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Combining Two Functions Let f and g be two functions with domains Df and Dg , respectively. Then we can form new functions by adding, subtracting, multiplying or dividing. These new functions, f + g , f − g , fg and f /g , are defined in the usual way: (f − g )(x) = f (x) − g (x) f f (x) (x) = g g (x) (f + g )(x) = f (x) + g (x) (fg )(x) = f (x)g (x) The domains of f + g , f − g and fg are the same and are equal to the intersection: Df ∩ Dg . That is, everything that is in common to both the domain of f and the domain of g . Since division by zero is not allowed, the domain of f /g is {x ∈ Df ∩ Dg : g (x) 6= 0}. That is, everything that is in common to both the domain of f and the domain of g , except the points that make g (x) equal to zero. 6 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Performing Function Operations OPTIONAL Example—if not covered, try it on your own and check with the ACW Let f (x) = x 2 + 3 and g (x) = x − 2. Find f + g , f − g , fg , and f /g . Solution: We have (f + g )(x) = (f − g )(x) = (fg )(x) = f (x) g = 7 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Composition Let f and g be two functions with domains Df and Dg , respectively. The composition of f and g , denoted by f ◦ g , is defined as: (f ◦ g )(x) = f (g (x)). The domain of f ◦ g is {x ∈ Dg : g (x) ∈ Df }. That is, it contains all values x in the domain of g such that g (x) is in the domain of f . Example Let f (x) = x 2 and g (x) = √ x. Find the domain of f ◦ g . Solution: 8 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Performing Function Composition OPTIONAL Example—if not covered, try it on your own and check with the ACW Let f (x) = x 2 + 3 and g (x) = x − 2. Find f ◦ g and g ◦ f . Solution: We have (f ◦ g )(x) = (g ◦ f )(x) = Observation: In general, 9 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Demand and Supply Functions A demand function p = d(x) with unit price p and x units produced is characterized as a decreasing function. A supply function p = s(x) with p and x as above is characterized as an increasing function. The equilibrium point (x0 , p0 ) provides the quantity x and price p at which the demand equals the supply. p p0 (x0 , p0 ) x0 x 10 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Finding the Equilibrium Point Example The demand function for a certain commercial product is given by p = d(x) = −0.01x 2 − 0.2x + 8 and the corresponding supply function is given by p = s(x) = 0.01x 2 + 0.1x + 3 where p is expressed in dollars and x is measured in units of a hundred. Find the market equilibrium. Solution: We need to solve the system of equations by equating the demand and supply functions. 11 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Cost, Revenue and Profit Functions The linear cost function C is given by C (x) = mx + b, where m is the variable cost per unit, b is the overall fixed cost, and x is the number of units produced. The average cost function C is given by C (x) , x where C (x) is the cost function. The revenue function R is given by R(x) = px = xf (x), where p is the unit price, x is the number of units demanded, and f is the demand function. The profit function P is given by P(x) = R(x) − C (x), where C (x) is the cost function and R(x) is the revenue function. The break-even point (x, p) provides the quantity x and price p at which the revenue equals cost. Example A certain firm determines that the total cost C (x) in dollars of producing and selling x units is given by C (x) = 20x + 100. Management plans to charge $24 per unit. (a) How many units must be sold for the firm to break even? (b) What is the profit if 100 units are sold? (c) How many units must be sold to produce a profit of $900? 12 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice SIDE NOTES: Symmetry, Transformations and Compositions Finding the Break-even Point Solutions: y ($) 1000 y = R(x) y = C (x) profit 800 600 400 break-even loss 200 x (units) 10 20 30 40 50 13 LEGEND: ! pay attention for exam underline key concepts ? confusing, read course notes → need to practice