Chapter 2: Analysis of Graphs of Functions


Chapter 2: Analysis of Graphs of Functions

2.1 Graphs of Basic Functions and Relations; Symmetry

2.2 Vertical and Horizontal Shifts of Graphs

2.3 Stretching, Shrinking, and Reflecting Graphs

2.4 Absolute Value Functions: Graphs, Equations, Inequalities, and Applications

2.5 Piecewise-Defined Functions

2.6 Operations and Composition


2.6 Operations and Composition

• The domain of is the intersection of the domains of f and g, while the domain of f /g is the intersection of the domains of f and g for which

Operations on Functions

Given two functions f and g, then for all values of x for which both and are defined, the functions are defined as follows.

)(xf)(xg gfgfgfgf / and , , ,

0)(,)()()(

)()())(()()())(()()())((

xgxgxfx

gf

xgxfxgfxgxfxgfxgxfxgfSum

DifferenceProduct

Quotient

gfgfgf and , ,

.0)( xg


2.6 Examples Using Operations on Functions

Analytic Solution (a)

(b)

(c)

(d)

)0((d))5)(((c))3)(((b))1)(((a)

following. theofeach Find .53)( and 1)(Let 2

gfgfgfgf

xxgxxf

1082)1()1()1)(( gfgf

14)4(10)3()3()3)(( gfgf

5202026)5()5()5)(( gfgf

51

)0()0()0(

gf

gf


2.6 Graphing Calculator Capabilities of Function Notation

• We can support the analytic solution of the previous example with the calculator by using its function notation capability.

• Enter f as and g as 1y .

2y


2.6 Examples Using Operations on Functions

Solution (a)

(b)

(c)

).( ofdomain The)()((b)))(((a)

following. theofeach Find .12)( and 98)(Let

xgfcx

gfxgf

xxgxxf

1298)()())(( xxxgxfxgf

1298

)()()(

x

xxgxfx

gf

.,21 is g / ofdomain the

Therefore, domain. thefrom 21 exclude we,02

1 Since

.,21 interval get the we,inequality thisSolving

.012such that for numbers all : ofDomain numbers real all : ofDomain

f

g

xxgf


If possible, use the given graph of f and g to evaluate(a) (b) (c)

Solution(a)

(b)

(c)

2.6 Evaluating Combinations of Functions

undefined. is )2)(( thereforeundefined, is )2( and 3)2( gfgf

)2)(( gf)1)(( gf )0)(/( gf

.3)1(3)1()1()1)(( so ,1)1(,3)1( gfgfgf

undefined. is )0()0()0)(/(but ,0)0(,1)0(

gfgfgf


2.6 The Difference Quotient

The expression is called the difference quotient.

Figure 67 pg 2-153

hxfhxf

xhxxfhxfm )()(

)()()( :linesecant of Slope

hxfhxf )()(


2.6 Looking Ahead to Calculus

• The difference quotient is essential in the definition of the derivative of a function.– the slope of the secant line is an average rate of change

• The derivative is used to find the slope of the tangent line to the graph of a function at a point.– the slope of the tangent line is an instantaneous rate of change

• The derivative is found by letting h approach zero in the difference quotient.– i.e. the slope of the secant line approaches the slope of the tangent

line as h gets close to zero


2.6 Finding the Difference Quotient

Let Find the difference quotient and simplify.

Solution

.32)( 2 xxxf

hxxhxhx

hxfhxf

xfhxf )()(

)32()(3)(2)()( 22

hxxhxhxhx 3233)2(2 222

hxxhxhxhx 3233242 222

hhhxh 324 2

hhxh )324(

324 hx


2.6 Composition of Functions

Composition of Functions

If f and g are functions, then the composite function, or composition, of g and f is

for all x in the domain of f such that is in the domain of g.

)]([))(( xfgxfg

)(xf


2.6 Application of Composition of Functions

• Suppose an oil well off the California coast is leaking.

– Leak spreads in circular layer over water– Area of the circle is

• At any time t, in minutes, the radius increases 5 feet every minute.

– Radius of the circular oil slick is • Express the area as a function of time using

substitution.

2)( rrA

ttr 5)(

2

2

25)5(]5[)]([

tttAtrA


2.6 Evaluating Composite Functions

ExampleGiven find (a) and (b)

Solution(a)

(b)

,1

4)( and 12)(

x

xgxxf )2)(( gf ).3)(( fg

)]2([)2)(( gfgf 414

124)2(

g

)4(f 7181)4(2)4( f

7

)]3([)3)(( fgfg 7161)3(2)3( f

)7(g21

84

174)7(

g

21


2.6 Finding Composite Functions

Let and Find (a) and (b)

Solution(a)

(b)

Note:

14)( xxf .52)( 2 xxxg ))(( xfg ).)(( xgf

)14()]([))(( xgxfgxfg

12081)52(4

)52(

2

2

2

xxxxxxf

)14(5)14(2 2 xx520)1816(2 2 xxx

52021632 2 xxx73632 2 xx

)]([))(( xgfxgf

is not always equal to . When they are, it is a special case.g f f g


2.6 Finding Functions that Form a Composite Function

Suppose that Find f and g so that

SolutionNote the repeated quantity Let

Note that there are other pairs of f and g that also work.

.3)5(4)5()( 232 xxxh).())(( xhxgf

.52 x then,34)( and 5)( 32 xxxfxxg

)(3)5(4)5(

)5(232

2

xhxx

xf

)]([))(( xgfxgf


2.6 Application of Composite Functions

Finding and Analyzing Cost, Revenue, and Profit Suppose that a businessman invests $1500 as his fixed cost in a new venture that produces and sells a device that makes programming a VCR easier. Each device costs $100 to manufacture.

(a) Write a linear cost function with x equal to the quantity produced.(b) Find the revenue function if each device sells for $125.(c) Give the profit function for the item.(d) How many items must be sold before the company makes a profit?(e) Support the result with a graphing calculator.



Solution(a) Using the slope-intercept form of a line, let

(b) Revenue is price quantity, so

(c) Profit = Revenue – Cost

(d) Profit must be greater than zero

or ,bmxC(x)

.1500100)( xxC

.125)( xpxxR

)()()( xCxRxP

.150025)( tosimplifies )1500100(125 xxPxx

profit. a make tosold bemust devices 61least At .60get wefor solving and ,0150025 xxx



(e) Let ).( and ),(),( 12321 xPyyyxRyxCy

Figure 71a pg 2-162


2.6 Applying a Difference of Functions

Example The surface area of a sphere S with radius r is S = 4 r2.

(a) Find S(r) that describes the surface area gained when r increases by 2 inches.

(b) Determine the amount of extra material needed to manufacture a ball of radius 22 inches as compared to a ball of radius 20 inches.

22 4)2(4)( rrrS

inches square extra 1056 336

204)220(4)20( 22

S

Chapter 2: Analysis of Graphs of Functions

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Transcript of Chapter 2: Analysis of Graphs of Functions