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157SN03-functions

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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
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the Creative Commons license (CC BY-NC-SA), allowing anyone to
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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
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underline key concepts
? confusing, read course notes
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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
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? 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
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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
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→ 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
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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
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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
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! 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
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? 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
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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
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? 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
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? 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 $900?
12
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SIDE NOTES:
Symmetry, Transformations and Compositions
Finding the Break-even Point
Solutions:
y ($)
1000
y = R(x)
y = C (x)
profit
800
600
400
break-even
loss
200
x (units)
10
20
30
40
50
13
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underline key concepts
? confusing, read course notes
→ need to practice
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