Lesson 2: A Catalog of Essential Functions (slides)

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Sec on 2.2A Catalogue of Essen al Func ons

V63.0121.011, Calculus IProfessor Ma hew Leingang

New York University

Announcements

I First WebAssign-ments are due January 31I First wri en assignment is due February 2I First recita ons are February 3

Announcements

I First WebAssign-mentsare due January 31

I First wri en assignmentis due February 2

I First recita ons areFebruary 3

ObjectivesI Iden fy different classes of algebraicfunc ons, including polynomial(linear,quadra c,cubic, etc.), ra onal,power, trigonometric, and exponen alfunc ons.

I Understand the effect of algebraictransforma ons on the graph of afunc on.

I Understand and compute thecomposi on of two func ons.

Recall: What is a function?

Defini onA func on f is a rela on which assigns to to every element x in a setD a single element f(x) in a set E.

I The set D is called the domain of f.I The set E is called the target of f.I The set { y | y = f(x) for some x } is called the range of f.

Four ways to represent a function

I verbally—by a descrip on in wordsI numerically—by a table of values or a list of dataI visually—by a graphI symbolically or algebraically—by a formula

Today the focus is on the different kinds of formulas that can beused to represent func ons.

Classes of Functions

I linear func ons, defined by slope and intercept, two points, orpoint and slope.

I quadra c func ons, cubic func ons, power func ons,polynomials

I ra onal func onsI trigonometric func onsI exponen al/logarithmic func ons

OutlineAlgebraic Func ons

Linear func onsOther polynomial func onsOther power func onsGeneral ra onal func ons

Transcendental Func onsTrigonometric func onsExponen al and logarithmic func ons

Transforma ons of Func ons

Composi ons of Func ons

Linear functionsLinear func ons have aconstant rate of growth andare of the form

f(x) = mx+ b.

The slopem represents the“steepness” of the graphedline, and the intercept brepresents an ini al value ofthe func on.

.. x.

y

..

(0, b)

..

(x1, y1)

..

(x2, y2)

.

∆x = x2 − x1

.

∆y = y2 − y1

.

m =∆y∆x

Modeling with Linear FunctionsExample

Assume that a taxi costs $2.50 to get in and $0.40 per 1/5 mile. Writethe fare f(x) as a func on of distance x traveled.

AnswerThe ini al fare is $2.50, and the change in fare per mile is$0.40/0.2mi = $2/mi. So if x is in miles and f(x) in dollars, theequa on is

f(x) = 2.5+ 2x

Modeling with Linear FunctionsExample

Assume that a taxi costs $2.50 to get in and $0.40 per 1/5 mile. Writethe fare f(x) as a func on of distance x traveled.

AnswerThe ini al fare is $2.50, and the change in fare per mile is$0.40/0.2mi = $2/mi. So if x is in miles and f(x) in dollars, theequa on is

f(x) = 2.5+ 2x

A Biological ExampleExample

Biologists have no ced that the chirping rate of crickets of a certainspecies is related to temperature, and the rela onship appears to bevery nearly linear. A cricket produces 113 chirps per minute at 70 ◦Fand 173 chirps per minute at 80 ◦F.(a) Write a linear equa on that models the temperature T as a

func on of the number of chirps per minute N.(b) If the crickets are chirping at 150 chirps per minute, es mate the

temperature.

Biological Example: SolutionSolu on

I The point-slope form of the equa on for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equa on y− y0 = m(x− x0).

I The slope of our line is80− 70

173− 113=

1060

=16

I So an equa on rela ng T and N is

T− 70 =16(N− 113) =⇒ T =

16N− 113

6+ 70

Biological Example: SolutionSolu on

I The point-slope form of the equa on for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equa on y− y0 = m(x− x0).

I The slope of our line is80− 70

173− 113=

1060

=16

I So an equa on rela ng T and N is

T− 70 =16(N− 113) =⇒ T =

16N− 113

6+ 70

Biological Example: SolutionSolu on

I The point-slope form of the equa on for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equa on y− y0 = m(x− x0).

I The slope of our line is80− 70

173− 113=

1060

=16

I So an equa on rela ng T and N is

T− 70 =16(N− 113) =⇒ T =

16N− 113

6+ 70

Solution continued

I So an equa on rela ng T and N is

T− 70 =16(N− 113) =⇒ T =

16N− 113

6+ 70

I If N = 150, then

T =376

+ 70 = 7616◦F

Solution continued

I So an equa on rela ng T and N is

T− 70 =16(N− 113) =⇒ T =

16N− 113

6+ 70

I If N = 150, then

T =376

+ 70 = 7616◦F

Other polynomial functionsI Quadra c func ons take the form

f(x) = ax2 + bx+ c

The graph is a parabola which opens upward if a > 0,downward if a < 0.

I Cubic func ons take the form

f(x) = ax3 + bx2 + cx+ d

Other polynomial functionsI Quadra c func ons take the form

f(x) = ax2 + bx+ c

The graph is a parabola which opens upward if a > 0,downward if a < 0.

I Cubic func ons take the form

f(x) = ax3 + bx2 + cx+ d

Other power functions

I Whole number powers: f(x) = xn.

I nega ve powers are reciprocals: x−3 =1x3.

I frac onal powers are roots: x1/3 = 3√x.

General rational functions

Defini onA ra onal func on is a quo ent of polynomials.

Example

The func on f(x) =x3(x+ 3)

(x+ 2)(x− 1)is ra onal.

I The domain is all real numbers except−2 and 1.I The func on is 0 when x = 0 or x = −3.

OutlineAlgebraic Func ons

Linear func onsOther polynomial func onsOther power func onsGeneral ra onal func ons

Transcendental Func onsTrigonometric func onsExponen al and logarithmic func ons

Transforma ons of Func ons

Composi ons of Func ons

Trigonometric functions

I Sine and cosineI Tangent and cotangentI Secant and cosecant

GeoGebra applets to graph these

Exponential and logarithmicfunctions

I exponen al func ons (for example f(x) = 2x)I logarithmic func ons are their inverses (for example

f(x) = log2(x))

GeoGebra applets to graph these

OutlineAlgebraic Func ons

Linear func onsOther polynomial func onsOther power func onsGeneral ra onal func ons

Transcendental Func onsTrigonometric func onsExponen al and logarithmic func ons

Transforma ons of Func ons

Composi ons of Func ons

Transformations of FunctionsTake the squaring func on and graph these transforma ons:

I y = (x+ 1)2

I y = (x− 1)2

I y = x2 + 1I y = x2 − 1

Observe that if the fiddling occurs within the func on, atransforma on is applied on the x-axis. A er the func on, to they-axis.

Transformations of FunctionsTake the squaring func on and graph these transforma ons:

I y = (x+ 1)2

I y = (x− 1)2

I y = x2 + 1I y = x2 − 1

Observe that if the fiddling occurs within the func on, atransforma on is applied on the x-axis. A er the func on, to they-axis.

Vertical and Horizontal ShiftsSuppose c > 0. To obtain the graph of

I y = f(x) + c, shi the graph of y = f(x) a distance c units . . .

upward

I y = f(x)− c, shi the graph of y = f(x) a distance c units . . .

downward

I y = f(x− c), shi the graph of y = f(x) a distance c units . . .

tothe right

I y = f(x+ c), shi the graph of y = f(x) a distance c units . . .

tothe le

Vertical and Horizontal ShiftsSuppose c > 0. To obtain the graph of

I y = f(x) + c, shi the graph of y = f(x) a distance c units . . .upward

I y = f(x)− c, shi the graph of y = f(x) a distance c units . . .

downward

I y = f(x− c), shi the graph of y = f(x) a distance c units . . .

tothe right

I y = f(x+ c), shi the graph of y = f(x) a distance c units . . .

tothe le

Vertical and Horizontal ShiftsSuppose c > 0. To obtain the graph of

I y = f(x) + c, shi the graph of y = f(x) a distance c units . . .upward

I y = f(x)− c, shi the graph of y = f(x) a distance c units . . .downward

I y = f(x− c), shi the graph of y = f(x) a distance c units . . .

tothe right

I y = f(x+ c), shi the graph of y = f(x) a distance c units . . .

tothe le

Vertical and Horizontal ShiftsSuppose c > 0. To obtain the graph of

I y = f(x) + c, shi the graph of y = f(x) a distance c units . . .upward

I y = f(x)− c, shi the graph of y = f(x) a distance c units . . .downward

I y = f(x− c), shi the graph of y = f(x) a distance c units . . . tothe right

I y = f(x+ c), shi the graph of y = f(x) a distance c units . . .

tothe le

Vertical and Horizontal ShiftsSuppose c > 0. To obtain the graph of

I y = f(x) + c, shi the graph of y = f(x) a distance c units . . .upward

I y = f(x)− c, shi the graph of y = f(x) a distance c units . . .downward

I y = f(x− c), shi the graph of y = f(x) a distance c units . . . tothe right

I y = f(x+ c), shi the graph of y = f(x) a distance c units . . . tothe le

Why?

Ques on

Why is the graph of g(x) = f(x+ c) a shi of the graph of f(x) to thele by c?

AnswerThink about x as me. Then x+ c is the me c into the future. Torec fy the future of the graph of f with that of g, pull the graph of f cinto the past.

Why?

Ques on

Why is the graph of g(x) = f(x+ c) a shi of the graph of f(x) to thele by c?

AnswerThink about x as me. Then x+ c is the me c into the future. Torec fy the future of the graph of f with that of g, pull the graph of f cinto the past.

Illustrating the shift

I Adding cmoves x to theright

I But then f is appliedI To get the graph of

f(x+ c), the valuef(x+ c)must be above x

I So we translate backward

...x

..

(x, f(x))

..x+ c

..

(x+ c, f(x+ c))

..

(x, f(x+ c))

Illustrating the shift

I Adding cmoves x to theright

I But then f is appliedI To get the graph of

f(x+ c), the valuef(x+ c)must be above x

I So we translate backward

...x

..

(x, f(x))

..x+ c

..

(x+ c, f(x+ c))

..

(x, f(x+ c))

Illustrating the shift

I Adding cmoves x to theright

I But then f is applied

I To get the graph off(x+ c), the valuef(x+ c)must be above x

I So we translate backward

...x

..

(x, f(x))

..x+ c

..

(x+ c, f(x+ c))

..

(x, f(x+ c))

Illustrating the shift

I Adding cmoves x to theright

I But then f is appliedI To get the graph of

f(x+ c), the valuef(x+ c)must be above x

I So we translate backward

...x

..

(x, f(x))

..x+ c

..

(x+ c, f(x+ c))

..

(x, f(x+ c))

Illustrating the shift

I Adding cmoves x to theright

I But then f is appliedI To get the graph of

f(x+ c), the valuef(x+ c)must be above x

I So we translate backward ...x

..

(x, f(x))

..x+ c

..

(x+ c, f(x+ c))

..

(x, f(x+ c))

Now try these

I y = sin (2x)I y = 2 sin (x)I y = e−x

I y = −ex

Scaling and flippingc < 0 c > 0

|c| > 1 |c| < 1 |c| < 1 |c| > 1

f(cx) . . . .

H compress, flip H stretch, flip H stretch H compress

cf(x) . . . .

V stretch, flip V compress, flip V compress V stretch

OutlineAlgebraic Func ons

Linear func onsOther polynomial func onsOther power func onsGeneral ra onal func ons

Transcendental Func onsTrigonometric func onsExponen al and logarithmic func ons

Transforma ons of Func ons

Composi ons of Func ons

Composition of FunctionsCompounding in Succession

..f . g.

g ◦ f

.x . (g ◦ f)(x).f(x).

Composing

Example

Let f(x) = x2 and g(x) = sin x. Compute f ◦ g and g ◦ f.

Solu on

I (f ◦ g)(x) = sin2 xI (g ◦ f)(x) = sin(x2)

Note they are not the same.

Composing

Example

Let f(x) = x2 and g(x) = sin x. Compute f ◦ g and g ◦ f.

Solu on

I (f ◦ g)(x) = sin2 xI (g ◦ f)(x) = sin(x2)

Note they are not the same.

Decomposing

Example

Express√

x2 − 4 as a composi on of two func ons. What is itsdomain?

Solu onWe can write the expression as f ◦ g, where f(u) =

√u and

g(x) = x2 − 4. The range of g needs to be within the domain of f. Toinsure that x2 − 4 ≥ 0, we must have x ≤ −2 or x ≥ 2.

Decomposing

Example

Express√

x2 − 4 as a composi on of two func ons. What is itsdomain?

Solu onWe can write the expression as f ◦ g, where f(u) =

√u and

g(x) = x2 − 4. The range of g needs to be within the domain of f. Toinsure that x2 − 4 ≥ 0, we must have x ≤ −2 or x ≥ 2.

Summary

I There are many classes of algebraic func onsI Algebraic rules can be used to sketch graphs