Calculus Module - University of Pennsylvania

1

GSE/TFA Calculus Module

Calculus ModuleSaturday, March 8, 2008

Nakia Rimmer M.S., M.A.


Top Ten Pre-calculus skills needed to perform well in Calculus

10. Knowledge of Linear Functions

9. Knowledge of Polynomial Functions

8. Knowledge of Exponential Functions

7. Knowledge of Logarithmic Functions

6. Knowledge of Trigonometric Functions


Top Ten Pre-calculus skills needed to perform well in Calculus

5. Use of Technology

(TI graphing Calc. / Maple / Internet)

4. Factoring

3. Ability to Simplify Fractions

2. Simplifying

1. Avoid Common Algebra Mistakes


10. Knowledge of Linear FunctionsLinear Functions

Slope

risem

run=

change in ym

change in x=

As you move from one point

on the line to another

To gain a general

understanding, we assume

we move from left to right.

rise + run +( )0 m m is positive>

rise− run +( )0 m m is negative<

0rise = run +( )0 m horizontal line=

or rise = + −0run =

( ). m undef vertical line=


( ) ( )1 1 2 2Given 2 points: , ,x y x y

Find the slope of the line through the points.

2 1

2 1

y ym

x x

−=

−

( ) ( )7, 4 , 7,3− −

( )3 4

7 7m

− −=

− −7

14m =

−

1

2m = −

Parallel Lines

Perpendicular Lines

have

same slope

have

opposite sign

reciprocal

slopes



10. Knowledge of Linear FunctionsEquation of a line

Slope Intercept Form

y mx b= +

Point - Slope Form

( )1 1y y m x x− = −

General Form

0Ax By C+ + =

Horizontal Liney is constant,

x can be anything

y b=

Vertical Linex is constant,

y can be anythingx a=

In order to find the equation of the line you need:

a) a point on the line

b) the slope of the line

Am

B= −

2



a) Given the slope and a point that the lines goes through

Find the equation of a line:

2 3 5x y− =2

3

Am

B= − = −

−2

3m =

3

2m = −

( )3

7 42

y x− = − −( ). 4,7pt

( )1 1y y m x x− = −

37 6

2y x− = − +

313

2y x= − + B GSE/TFA Calculus Module

b) Given 2 points on the line

Find the equation of a line:

( )7,64Q ( )9.5,84K

84 64

9.5 7m

−=

−

208

2.5= =

( )64 8 7y x− = −

( )1 1y y m x x− = −

64 8 56y x− = −

8 8y x= +

C



9. Knowledge of Polynomial Functions

2

3

Parabola or Quadratic

y x= −

3 26 5 12

Cubic

y x x x= + + −

4 3 213 12

Quartic

y x x x x= − − + −5 4 3 25 15 85 26 120

Quintic

y x x x x x= − − + − −GSE/TFA Calculus Module

9. Knowledge of Polynomial FunctionsQuestions about polynomial functions that we need to answer :

�What is the domain and range of the polynomial?

Possible x-values Possible y-values

�What are the roots of the polynomial?

Set it = to 0 and solve for x.

�What are the local maximum (minimum) values of the polynomial

and for what values is the function increasing or decreasing?

What happens to the function as x gets large

or as x get small ?( )x → ∞

( )x → −∞

�What degree is it?

Degree = highest exponent on x

We’ll need calculus to answer this question for polynomials

of degree .3≥

�What is the end behavior of the polynomial?



Exponential Functions

xy a=

4xy =

3xy =

2xy =

x -2 -1 0 1 2 3

1 2 4 8

1 4 16 64

2xy =

4xy =

1/ 4

1/16

1/ 2

1/ 4

Domain: ( ),−∞ ∞

Range: ( )0,∞

Intercept: ( )0,1

Horizontal Asymptote: ( )0 y x axis= −

Increasing

Continuous

For a > 1



3xy =

3xy = −

3xy =3 xy −=

3xy =23xy += 3xy =

x axis flip−

y axis flip− Shift up

Shift left

3 2xy = +23xy +=

3



e – The Natural Base

2.7182818284e ≈ …

2 3e< <

xy e=

Like , the decimal expansion of

never terminates and never repeats.

It is an irrational number.

π e

When we get to calculus we'll see that

is the only function that has itself

as its derivative.

xy e=

e is the base to the natural logarithm function


Logarithmic Functions

log y

ay x x a= ⇔ =

base

exponent

resultexponentbase

result62 64=�

2log 64 = 6

10log log=If ever you see log

without a base then

it is understood

that the base is 10

ln loge=The natural

logarithm ln has e

as its base

Your calculator can do base 10 and base e, but to get other bases

you’ll have to use the change of base formula.

7. Knowledge of Logarithmic Functions


7. Knowledge of Logarithmic FunctionsThe Natural Log Function

and ln are inverse functionsxe x

xy e= lny x=

y x=0 1e = ⇔ log 1 0e =

�ln1 0=

1e e= ⇔ log 1e e =

�ln 1e =

lny x=

Domain:

( ),−∞ ∞Range:

( )0,∞

Intercept: ( )1,0

Vertical Asymptote: ( )0 x y axis= −

Increasing Continuous



45°

45°

1

1

260°

30°

1

3

Special Right Triangles

2

Oppositesin

Hypotenusex =

Adjacentcos

Hypotenusex =

Oppositetan

Adjacentx =



( )cos ,sinx x



http://www.math.udel.edu/~rimmer/math222/graphs.html

For animated graphs of the sine and cosine functions go to :

siny x= in radiansx

( )cosy x=

4


5. Use of Technology

TI Graphing Calculator

http://mathforum.org/mathtools/activity/20091/

http://tutorial.math.lamar.edu/AllBrowsers/2413/2413.asp

Maple

Internet

Virtual TI – Free PC Program that emulates any version of TI

graphing calculator.

Computer Algebra System being integrated into the

curriculum of most universities. Great for graphing

especially in 3d.


4. Factoring

( ) 3 2Find the roots of 12 66 72f x x x x= − +

3 212 66 72 0x x x− + =

( )26 2 11 12 0x x x− + =

( )( )6 2 3 4 0x x x− − =

6 0 2 3 0 4 0x x x= − = − =

3

20 4x xx= = =


3. Ability to Simplify Fractions

Simplify

xy

cx

x

=

+

x xy

c xx

x

= ⋅

+

2

2

x x xy y

c x x cx

x

= ⋅ ⇒ =+

+

( )( )

Solve

2 1 1

1 2 2 1x x x x= −

− − − −

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

( )( )2 1 1

2 1 2 1 2 11 2 2 1

x x x x x xx x x x

⋅ − − = ⋅ − − − ⋅ − −− − − −

( ) ( )2 2 1 1x x− = − − ⇒ 2 4 2x x− = − ⇒ 2x =

2 and 1 div. by 0x x≠ ≠ ⇒

but 2 sox ≠ No Solution GSE/TFA Calculus Module

2. Simplifying

I have __________, the back of the book has __________.

“Back of the book Drama”51

2 21 3

2 2x x

−−

+2

1 31

2 xx

+

512 2

1 3

2 2x x

−−

+ 52

1 3

22 xx= +

1 42 2

1 1 3

22 x xx= +

2

1 31

2 xx

= +


1. Avoid Common Algebra Mistakes

http://www.mathmistakes.info/mistakes/algebra/index.html

A great website discussing this topic is :

( )2 2 2x y x y+ ≠ +

x y x y+ ≠ +

( )2 23 3− ≠ −

( )

( )

3 5

3 5

f x x

f x h x h

= −

+ ≠ − +


5


Limits

Notation

lim ( )x a

f x L→

=

( )read as : "the limit of , as approaches , equals "f x x a L

( )if we can make the values of as close to as wef x L

like by taking to be sufficiently close to x a

(on either side of ) but not equal to .a a

( ) as f x L x a→ →


2

23

9.1 lim

4 3x

xex

x x→

−

− +

( ) ( )

( )( )

2

23 3

3 39lim lim

4 3 3 1x x

x xx

x x x x→ →

− +−=

− + − −

( )3

3limx

x

→

−=

( )

( )

3

3

x

x

+

− ( )1x −

3

3lim

1x

x

x→

+=

−3=

Limits


Limits

22

1 4. 2 lim

2 4xex

x x→−

+

+ −

( )

( )

( ) ( ) ( )22 2

21 4 1 4lim lim

2 4 2 2 2 2x x

x

x x x x x x→− →−

− + = ⋅ + + − + − − +

( )( )2

2 4lim

2 2x

x

x x→−

− +=

− +

( )( )2

2lim

2 2x

x

x x→−

+=

− +

2

2limx

x

→−

+=

( ) ( )2 2x x− +

( )2

1lim

2x x→−=

−

1

4= − GSE/TFA Calculus Module

Definition of the derivative

( )( ) ( )

0limh

f x h f xf x

h→

+ −′ =

• Slope of the tangent line to a curve at a pt.

• Function that measures rate of change


( )

( )

Let 3 5 .

a) Use the definition of the derivative to find .

f x x

f x

= −

′

( )( ) ( )

0limh

f x h f xf x

h→

+ −′ =

( )( )

0

3 5 3 5limh

x h xf x

h→

− + − −′ =

( )( )( ) ( )( )

( )0

3 5 3 5 3 5 3 5lim

3 5 3 5h

x h x x h xf x

h x h x→

− + − − − + + −′ = ⋅

− + + −

( )( ) ( )

( )( )0

3 5 3 5lim

3 5 3 5h

x h xf x

h x h x→

− + − −′ =

− + + −

( )0

3limh

f x→

′ =5x− 5 3h− − 5x+

( )( )3 5 3 5h x h x− + + −GSE/TFA Calculus Module

( )

( )

Let 3 5 .

a) Use the definition of the derivative to find .

f x x

f x

= −

′

( )0

5limh

hf x

→

−′ =

h ( )( )3 5 3 5x h x− + + −

( )0

5lim

3 5h

f x

x h→

−′ =

− +( )0 3 5x

+ −

( )( )

5

3 5 3 5f x

x x

−′ =

− + −

( )5

2 3 5f x

x

−′ =

−

6


( )

( )

Let 3 5 .

6b) Find the equation of the tangent line to at ,3

5

f x x

f x

= −

−

6

5

6 5

5 62 3 5

5

x

f

−= ⇒

− − ′ = −

−

6 5

5 2 3 6f

− − ′ =

+

6 5

5 6f m

− − ′ = =

6 5 with , 3, Find .

5 6y mx b x y m b

− −= + = = =

5 63 2

6 5b b

− − = + ⇒ =

52

6y x

−= +


Differentiation Rules

( ) ( ) ( )

( ) ( ) ( )

Sum/Difference Rule

h x f x g x

h x f x g x

= ±

′ ′ ′= ±

Let be a constant.

0

c

y c

y

=

′ =

( )

( )

y cf x

y cf x

=

′ ′=

( )

( ) 1

Power Rule

, is a real number n

n

f x x n

f x nx −

=

′ =

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

Product Rule

h x f x g x

h x f x g x f x g x

= ⋅

′ ′ ′= ⋅ + ⋅


Differentiation Rules

( ) ( )( )

( ) ( )( ) ( )

Chain Rule

(composition of functions)h x f g x

h x f g x g x

=

′ ′ ′= ⋅

( )( )

( )( )( )

( )( ) ( ) ( ) ( )

( )2

Quotient Rule

, 0f x

h x g xg x

g x f x f x g xh x

g x

= ≠

′ ′⋅ − ⋅′ =


( )

( )

Let 3 5 .

Use differentiation rules to find .

f x x

f x

= −

′

( ) ( )12 3 5 3 5f x x x= − = −

( ) ( ) ( )12

1 3 5 5

2f x x

−′ = − ⋅ −

( )5

2 3 5

f xx

−′ =

−

Same result as before with far less work!


More Derivatives

( )

( )

x

x

f x e

f x e

=

′ =

( ) ( )

( )

ln

1

f x x

f xx

=

′ =

( ) ( )

( ) ( )

sin

cos

f x x

f x x

=

′ =

( ) ( )

( ) ( )

cos

sin

f x x

f x x

=

′ = −


2 1( ) xy f x x e x

x= = + − +

Domain : 0x >

Range : 0y >

1x y= = 1e + 3.7183≈

4x y= =4 1

414e + 68.8482≈

1100

x y= = 0.01 99.9e + 100.9≈

( )Continuous on 0, ?∞ yes

7


2 1( ) xy f x x e x

x= = + − +


2 1( ) xy f x x e x

x= = + − +

2 1/ 2 1( ) xy f x x e x x−= = + − +

2

1 1( ) 2

2

xy f x x exx

′ ′= = + − −

1/ 2 21( ) 2 ( 1)

2

xy f x x e x x− −′ ′= = + − + −

Rewrite ( )f x

Take the derivative using the power rule

Simplify


What happens to the function ( ) when ( ) 0?f x f x′ <

What happens to the function ( ) when ( ) 0?f x f x′ >

( ) decreasesf x

( ) increasesf xWhat happens to the function ( ) when ( ) 0 and ( )

changes from decreasing to increasing?

f x f x f x′ =

( ) has a local minimumf xWhat happens to the function ( ) when ( ) 0 and ( )

changes from increasing to decreasing?

f x f x f x′ =

( ) has a local maximumf xGSE/TFA Calculus Module

( )f x

( )f x′

So we know that ( ) changes from negative to positive,

we just need to find out where ( ) 0.

f x

f x

′

′ =

2

1 1( ) 2 0

2

setxf x x e

xx′ = + − − =

This is too difficult to solve by hand so we try the graphing calculator.


5 6(3 4)y x−= +

Find y′

( )5 5 66(3 4) 15 Chain Rule and Power Ruley x x− −′ = + ⋅ −

( )2 4 7 ln( )y x x x= − +

( ) ( )2 12 4 ln( ) 4 7 Product Ruley x x x x

x

′ = − + − +

2

3

3 5 9

11 4 8

x xy

x x

+ −=

− +

( )( ) ( )( )

( )

3 2 2

23

11 4 8 6 5 3 5 9 33 4Quotient Rule

11 4 8

x x x x x xy

x x

− + + − + − −′ =

− +GSE/TFA Calculus Module

2A ship uses 5 dollars of fuel per hour when traveling

at a speed of miles per hour. The other expenses of

operating the ship amount to $ 2000 per hour. What

speed minimizes the cost of a 500-mil

x

x

e trip?

�

Let miles per hour be the speed.

Let ( ) be the cost in dollars.

$( ) and so

time

x

C x

dC x hour d r t t

hour r= × = ⋅ =

( )2 500 1,000,000( ) 5 2000 2,500C x x x

x x

= + = +

2

1,000,000( ) 2,500 0

set

C xx

′ = − =2 400 or 20x x⇒ = =

20 ( ) negative 20 ( ) positive local min.x C x x C x′ ′< ⇒ > ⇒ ⇒

20 miles per hour

(20) $100,000

x

C

=

=

8


A poster is to have area 125 square inches. The printed material is

to be surrounded by a margin of 3 inches at the top and 2 inches at

the bottom and sides. Find the dimensions of the poster that

maximize the area of the printed material.

3

2

2

2

x

y

x - 4

y - 5

125 sq. in.posterA xy= =

( )( )print 4 5

maximize this area

A x y= − −

Problem: It has 2 variables!

Solution: Use the fact that 125

125to give in terms of

xy

y x yx

=

⇒ =


2 2

500 5005 0 5

set

Ax x

′ = − + = ⇒ =

( )print

125 5004 5 125 5 20A x x

x x

= − − = − − +

is increasing is decreasing

10 0 and 10 0

We have a maximum.

A A

x A x A′ ′< ⇒ > > ⇒ <

⇒

��

print

500145 5A x

x= − −

2 100 10 in.xx == ⇒ 12.5 5

in12

.yyx

== ⇒


The antiderivative – the opposite of

taking a derivative of a function

( ) Find the function

whose derivative is ( ). Call it ( ).

f x dx

f x F x

⇒∫

� anyconstant

( ) ( )

This is called an indefinite integral .

f x dx F x C= +∫


The area under the graph of f (x),

above the x-axis between x = a and x = b

3

1

( ) 24.335 sq. units

This is called a definite integral .

f x dx ≈∫


If ( ) is continuous on [ , ] and

( ) is the antiderivative of ( ), then

f x a b

F x f x

( ) ( ) ( )b

a

f x dx F b F a= −∫


Evaluate the indefinite integral.

( )2 2 3x x dx− +∫3

2 33

xx x C= − + +

2 1x xdx

x

+ +∫ ( )3/ 2 1/ 2 1/ 2x x x dx−= + +∫

5/ 2 3/ 2 1/ 22 22

5 3x x x C= + + +

5/ 2 3/ 2 1/ 2

5 3 1

2 2 2

x x xC= + + +

2 1xx e x dxx

+ − +

∫ 3 3/ 21 2

ln( )3 3

xx e x x C= + − + +

9


Evaluate the definite integral.

4

1

2xdx

x

−∫

3

2

1

1xx e x dxx

+ − +

∫

( )1

3

0

5 4x x dx− +∫1

4 2

0

1 54

4 2x x x

= − +

1 5 1 10 16 74

4 2 4 4 41.75

4= − + = − + = =

16 10 26 248

3 3 3 3

2

3

− = − − = − =

( )4

1/ 2 1/ 2

1

2x x dx−

= −∫4

3/ 2 1/ 2

1

24

3x x

= −

( ) ( )3/ 2 1/ 22 24 4 4 4

3 3

= − − −

3

3 3/2

1

1 2ln( )

3 3

xx e x x

= + − +

( )3 3/22 1 29 3 ln(3) 24.33

35

3 3e e

= + − + − + − ≈


THANK YOU!

www.math.upenn.edu/~rimmer/tfa/tfa.htm

Calculus Module - University of Pennsylvania

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