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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.1 - Functions Math 103-991

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Chapter 1.1 - Functions • Relationships can be represented mathematically as function • An example of a function is f(x) = x2 + 4 where y=f(x) – X and Y are called variables – Because y depends on x, y is the dependent variable and x is the independent variable f(1)= (1) + 4 = 5

f(-2) = (−2) + 4 = 4 + 4 = 8

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Math 103 Class Notes

Sept 2-9, 2020

Piecewise-defined Functions • Functions can be defined by using more than one formula but only as a subset of the domain • An example of a piecewise-defined function is f(x) =

x +1 if x < 1 2x

Compute f(0)

if x ≥ 1

=0+1=1

Compute f(1) = 2(1) = 2 Compute f(3) = 2(3) = 6

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Chapter 1.1 - Domain • Unless specified otherwise, the domain of a function is the set of all numbers for which the function is defined – All the x values that will result in f(x) being defined • A polynomial such as f(x) = x3 + 2x +1 will have a domain of all real numbers (x€ R) • Domain must exclude all numbers that result in dividing by 0 or taking the square root of a negative number • Do domain examples from Section 1.1 Exercises in the text book 4

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.1 – Find Domain of given functions #19 g(x) =

𝒙𝟐 𝟓 𝒙𝟐

x+2 ≠ 0 x ≠ -2 Thus domain (-∞,-2)(-2,∞)

#21 f(x) = 𝟐𝒙 + 𝟔 2x+6 ≥ 0 2x ≥ -6 x ≥ -3 Thus domain [-3,∞) #23 f(t) = 9 - 𝑡 > 0 9 > 𝑡 ±3 > t

𝒕𝟐 𝟗𝒕𝟐

9-(−4) = -7

9-(0) = 9

-3

0

9-(4) = -7 3

Thus domain (-3,3) or -3 < t < 3

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Functions in Economics • Demand relates the unit price p in terms of numbers of units x – Example p(x) = -0.27x + 57 • Total Revenue = (number of items sold)(price per item); R(x) = x ∙ p(x) – Example R(x) = x (-0.27x + 57) = - 0.27x2 + 57x

• Cost is the total cost of producing x units – Example C(x) = 5x2 + 2.5x + 85

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Math 103 Class Notes

Sept 2-9, 2020

Functions in Economics • Revenue = - 0.27x2 + 57x Cost = 5x2 + 2.5x + 85 • Profit = Revenue – Cost = R(x) – C(x) = -0.27x2 + 57x – (5x2 + 2.5x + 85) = -0.27x2 + 57x – 5x2 - 2.5x - 85 = - 5.27𝑥  + 54.5x − 85

• Breakeven Point is when Revenue = Cost R(x) = C(x)

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Composition of Functions • Combining functions by substituting variables f(g(x))

• Do examples from Section 1.1 Exercises in the text book

Use FOIL

#25 Find f(g(x)) if f(u) =

3𝒖𝟐

= f(x+2) = 3(𝑥 + =

3(x2+2x+2x+4)

Find g(f(u))

+ 𝟐𝒖 − 𝟔 𝒂𝒏𝒅 𝒈 𝒙 = 𝒙 + 𝟐

2) +2

𝑥 + 2 − 6 = 3(x+2)(x+2)+2x + 4 - 6

+2x -2 = 3x2+6x+6x+12+2x-2 = 3x2+14x+10 x

First Outside Inside Last

+2

= g(3u2 + 2u -6) = 3u2 + 2u -6 +2 = 3u2 + 2u - 4

#45 Find f(x-1) where f(x) = (x+1)5 – 3x2 = (x-1+1)5 – 3(x-1)2 = x5- 3(x-1)(x-1) = x5- 3(x2-x-x+1) = x5- 3x2+ 6x -3

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.1 - Difference Quotient • Where f is a given function of x and h is a number • Do examples from Section 1.1 Exercises in the textbook

#33 Find the difference quotient of f(x) = 4 – 5x f(x+h)

=

𝒇 𝒙𝒉 𝒇 𝒙 𝒉

=

𝟒𝟓𝒙𝟓𝒉𝟒𝟓𝒙 𝒉

=

f(x)

𝟒𝟓 𝒙𝒉 [𝟒𝟓𝒙] 𝒉

=

 

= -5

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Chapter 1.1 - Difference Quotient #35 Find the difference quotient of f(x) = 4x- x2 𝟒(𝒙𝒉)(𝒙𝒉)𝟐 [𝟒𝒙𝒙𝟐 ] 𝟒𝒙𝟒𝒉 𝒙𝒉 𝒙𝒉 𝟒𝒙𝒙𝟐 = 𝒉 𝒉 𝟒𝒙𝟒𝒉(𝒙𝟐 𝟐𝒙𝒉𝒉𝟐 )𝟒𝒙𝒙𝟐 𝟒𝒙𝟒𝒉𝒙𝟐 𝟐𝒙𝒉 𝒉𝟐 𝟒𝒙𝒙𝟐 = = 𝒉 𝒉 𝟐 𝟒𝒉𝟐𝒙𝒉 𝒉

=

=

𝒉

= 4-2x-h

On UR Courses, I have suggested several odd questions to practice from this section in the textbook 10

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.2 – The Graph of the Function Math 103-991

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Chapter 1.2 – The Graph • To Plot points on a graph – Construct a table of chosen x values – Calculate f(x) = y – plot the coordinates on the graph & connect them

• Example: Graph y = -2x + 2 X XX

YY == -2x -2x +2 +2

-1-1

-2(-1) +2 + 2==22+2 +2=4 =4

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-2(0) + 2 = 0 + 2 = 2

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-2(1) + 2 = -2 + 2 = 0

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.2 – The Graph • Finding intercepts – X intercept is found by setting y =0 then solving for x – Y intercept is found by setting x =0 then solving for y

• Do examples from Section 1.2 Exercises #17 Sketch the graph of the function f(x) = 2x-1. Include all x and y intercepts. Find Y intercept: f(0) = 2(0) -1 = -1 Find X intercept: 0 = 2x-1 1 = 2x  

=x

To graph a straight line, you just need two points to plot  

Using the x and y intercept: (0,-1) and ( , 0)

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Chapter 1.2 – The Graph #25 Sketch the graph of the piece wise function f(x) = x -1 if x ≤ 0 x + 1 if x > 0 Plot points in the domain for each starting with x ≤ 0 xx

YY == xx -- 11

0

y = 0 - 1 = -1

-1

Y = - 1 - 1 = -2

Plot points for x > 0 x

Y=x+1

0

Y = 0 + 1 =1

1

Y=1+1=2

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.2 - Parabolas • Parabolas have a general equation of y = Ax2 + Bx +C as long as A ≠ 0 • All parabolas have a U shape • If A > 0, the parabola opens up and if A < 0, the parabola opens down • To find the x coordinate of the vertex use x =

 

• Do example #21 from Section 1.2 Exercises; graph f(x) = -x2 -2x +15 Find Y intercept: f(0) = -(0)2 -2(0) +15 = 15 Find X intercept: -x2 -2x +15 = 0 -(x2 +2x -15) = 0

Find vertex x =

 

-(x+5)(x-3 = 0 x = -5 and x = 3  () = = = −1 ()



Find y point of vertex by subbing x in for f(x); f(-1)= -(-1)2 -2(-1) +15 = -1 +2 +15 = 16 So vertex is (-1,16)

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Intersections of Graphs • The values of x for which 2 functions are equal are the x coordinates of the points where the graphs intersect • Steps to find the intersection of graphs – Solve each equation for y – Let the equations equal to each other – Solve for x to obtain the x coordinate – Substitute the x value into one of the equations to find the y coordinate

• Do examples from Section 1.2 Exercises

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Math 103 Class Notes

Sept 2-9, 2020

Intersections of Graphs #29 Find the points of intersection (if any) of the given pair of curves and draw the graphs; y = 3x + 5 and y = -x +3 Let the equations equal to each other

3x+5 = -x+3 Solve for x to obtain the x coordinate

4x = - 2  x= =-½ 

Substitute the x value into one of the equations to find the y coordinate

y(- ½) = - (- ½) + 3 = ½ +3 =

   

Thus Point of intersection: (- ½, ) Graph the point of intersection; note that 7/2 = 3.5 We need one more point on each graph, so find y intercepts and plot

For y = 3x+5, y(0) = 3(0)+5 = 5 For y = - x +3, y(0) = - 0 +3 = 3

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Intersections of Graphs #33 Find the points of intersection (if any) of the given pair of curves and draw the graphs; 3y – 2x = 5 and y + 3x = 9 First solve each equation for y

y = -3x + 9 3y = 2x + 5 y = 2/3 𝑥 + 5/3 Let the equations equal to each other  

𝑥+

 

= -3x + 9

Solve for x to obtain the x coordinate

2x + 5 = -9x +27 11x = 22 x=2 y(2) = -3 (2)+ 9 = 3 Thus Point of intersection: (2,3) We need one more point on each graph, so find y intercepts and plot

y(0) = -3(0)+9 = 9







y(0) =  (0) +  =  = 1.7

On UR Courses, I have suggested several odd questions to practice from this section in the textbook 18

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.3 – Linear Functions Math 103-991

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Chapter 1.3 – Linear Functions • The graph of a linear function is a straight line • Slope = rise/run = change in y / change in x = ∆y / ∆x • Do examples from Section 1.3 Exercises #3 Find the slope of the line passing through (2, 0) and (0, 2) m=

   

=

 



=  = −1

#5 Find the slope of the line passing through (2, 6) and (2, -4) m=

 

=

  

undefined

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.3 – Linear Functions • Vertical lines – Slope is undefined for any vertical line – Do example #13 from Section 1.3 Exercises – Sketch x = 3 No y intercept and x intercept of x = 3

• Horizontal lines – Slope is 0 for any horizontal line – Do example #14 from Section 1.3 Exercises – Sketch Y = 5 No x intercept and y intercept of y = 5 21

Chapter 1.3 – Linear Functions • Slope Intercept Form Y = mx +b where m = slope and b = y intercept #17 Find the slope and intercepts of 3x+2y=6, then graph the line Solve for y to obtain slope intercept form; 2y = -3x+6 y = -3/2 x +3 thus slope = m = -3/2 and y intercept = b =3 x intercept 3x + 2(0) = 6 3x = 6 So x = 2

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.3 – Linear Functions • Point Slope Form – Can be used to find the equation of the line when a point and a slope is known – Formula to use y – y0 = m (x-x0)

#27 Find the equation for the line through (1,0) and (0,1) Find slope m =

 

=

  

=

 

= −1

Use point slope formula: y - 0 = -1 (x-1) y = -1x +1 #25 Find the equation for the line through (2,5) and parallel to the x axis Slope =0 for a line that is parallel to the x axis, thus m = 0 Use point slope formula: y – 5 = 0 (x – 2) y=5 #26 Find the equation for the line through (2,5) and parallel to the y axis Slope is undefined for a line that is parallel to the y axis (vertical line) x=2

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Chapter 1.3 – Linear Functions Parallel lines have equal slopes #33 Find the equation for the line through (4, 1) and parallel to 2x+ y =3

Solve for y in the parallel line y = -2x +3 y= mx +b thus m = -2 use point slope formula: y - 1 = -2 (x-4) y = -2x +8 +1 y = -2x + 9

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Math 103 Class Notes

Sept 2-9, 2020

Chapter 1.3 – Linear Functions Perpendicular lines have slopes = -1 / m #35 Find the equation for the line through (3, 5) and perpendicular to x + y = 4 Solve for y in the perpendicular line y=-x+4  thus m =  = 1 use point slope formula: y - 5 = 1 (x-3) y = x - 3 +5 y=x+2

On UR Courses, I have suggested several odd questions to practice from this section in the textbook

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