C A L CABFOR THOSE WHO NEED A GOOD OVERVIEW BEFORE THE EXAM

by CHRIS REDMONDVitalApproach

WHERE IT COUNTSTM

A BRIEF GUIDE TO AP

2

Limits

&Continuity

What is a Limit?A limit is the height a function intends to reach at a given x value, whether or not it actually reaches it

We write:

lim f(x)=Lx c

ExamplEs

f(x)= x2-3x+2 lim f(x) = 6x 4A)

This is the substitution method. Simply plug the number you’re approaching (4) in for the variable (x).

This is the factoring method. Both the top and bottom of the fraction contain (x+3), so you can cancel those terms out and then use the substitution method.

B) f(x)= x2-9x+3

lim f(x) = 7x 10= (x+3)(x-3)

x+3

= (x-3)

When a Limit Does not exist

1.) f(x) approaches a different number from the right side of c and the left side at c

lim x 8

xx

lim x 0

1x

2.) f(x) increases/decreases without bound as x c

3

sCaLar muLtipLe

lim bf(x) = bLx c

Let b and c be real numbers. Let n be a positive integer. Let f and g be functions with the following limits:

lim f(x)=Lx c

lim g(x)=Kx c

power

lim [f(x)]n=Lnx c

produCt

lim [f(x)g(x)]=LKx c

ProPerties of Limits

theorem

sum/differenCe

x clim [f(x) g(x)]=L K+_ +_

lim b=bx c

lim x=cx c

lim xn=cnx c

Quotientf(x)g(x)

x c

LKlim =

lim sin x 0

1x

x c3.) f(x) oscillates between values as

4

If f and g are functions such that and thenlim g(x)=Lx c

lim f(x)=f(L)x L

lim f(g(x))=f(L)x c

If p is a polynomial function and c is a real number,

lim p(x)=p(c)x c

lim r(x)=x c

p(c)g(c)

If r is a rational function given by r(x)= and c is a real number such that g(c)=0 then,

p(x)g(x)

theorem

theorem

lim sin x = sin c x c

lim cot x = cot c

lim cos x = cos c lim sec x = sec c

lim tan x = tan c lim csc x = csc c

Let c be a real number and let f(x)=g(x) for all x=c in an open interval containing c. If exists, the also exists and =

lim g(x)x c

lim f(x)x c

lim g(x)x c

lim f(x)x c

theorem

theorem

x c

x c

x c

x c

x c

5

If h(x) f(x) g(x) for all x in an open interval containing c, except possibly at c itself, and if =L= , then lim h(x)

x clim g(x)x c

lim f(x)=Lx c

lim =1x 0

sin xx lim =0

x 0

1-cos xx lim =0

x 0

cos x-1x

A function is continuous on an open interval (a,b) if for every x in the interval, there is a y.

A function is defined at c if: 1) f(x) is defined 2) exists 3) = f(c)

lim f(x)x c

lim f(x)x c

removeabLecan be made continuous by defining or redefining f(c)

nonremoveabLecannot be made continuous by

defining or redefining f(c)

squeeze theorem

theorem

What is Continuity?

Continuity at a Point

DisContinuities

6

If f is continuous on [a,b] and k is any number between f(a) and f(b), then there is at least one number c in [a,b] such that f(c)=k

If r is a positive rational number and c is any real number then

lim = 0x

8

cxr

If xr is defined for x<0

lim = 0x - 8

cxr

1) lim x 2

x2-4x2 +4

2) lim x

8

4-x2

x2-1

3) lim x 3

x-3x2 -2x-3

intermeDiate VaLue theorem

Limits of infinity

PraCtiCe ProbLems

Find the following limits.

lim f(x)=L

7

8

derivatives

A derivative is the slope of a function.

At any x in the domain of the function y=f(x), the derivative is defined as

lim f(x+ x)-f(x)x

x 0OR lim y

xx 0

The function is said to be differentiable at every x for which this limit exists, and its derivative may be denoted by f’(x), y’, , or Dxy. Frequently x is replaced by h or some other symbol.

The derivative of y=f(x) at x=a, denoted by f’(a) or y’(a), may be defined as follows:

dydx

lim h 0

f(a+h)-f(a)hf’(a) =

dadx =0

dudx

ddx au=a

ddx xn=nxn-1 (Power Rule)

ddx

ddx

ddx (u v)= u v+ +

ddx

dudx

dvdx(uv)=u +v (Product Rule)

dudx

dvdxv -ud

dx( )= v=0 (Quotient Rule)uv

v2

What is a DeriVatiVe?

basiC DeriVatiVe formuLas

dudx

ddxsin u= cos u

9

_ _

dudx

ddxtan u=sec2u

dudx

ddxcot u= -csc2u

dudx

ddxsec u= sec u tan u

dudx

ddxcsc u= -csc u cot u

dudx

ddxeu=eu

dudx

ddxau=au ln a

dudx

ddxln u= 1

u

ddx [cf(x)]=c [f(x)]d

dx

ddx [f(g(x))=f”(g(x))g’(x)

ddx [u(x)]n=n[u(x)]n-1 du

dx

Constant muLtiPLe ruLe

Chain ruLe

GeneraL PoWer ruLe

dudx

ddxcos u= -sin u

10

A) ddx xy2 = x2y +y2 = 2xy +y2dy

dxdydx

B) ddx [x3y2] = x32y +3x2y2 = 2x3y +3x2y2dy

dxdydx

ddx [y3]=3y2 dy

dxUse chain rule (let y=f(x))

Implicit Differentiation comes into play when a functional relationship between x and y is defined by an equation of the form F(x, y)=0. For example:

x2+y2-9=0y2-4x=0

cos (xy)=y2-5

(f-1)’(b)=1

f’(a) (f-1)’(x)= 1f’(f-1(x))OR

The derivative of the inverse of a function at a point is the reciprocal of the derivative of the function at the corresponding point

imPLiCit Diferentiation

ExamplEs

DeriVatiVe of the inVerse of a funCtion

mean VaLue theorem

If the function f(x) is continuous at each point on the closed interval a x b and has a derivative at each point on the open interval a x b, then there is at least one number c,a c b, such that =f’(c). This important theorem relates average rate of change and instantaneous rate of change.

f(b)-f(a)b-a

11

Let f be defined at c. If f’(c)=0 or if f is not differentiable at c, then c is a critical number.

If f is continuous on a closed interval [a,b] then f has both a max and a min in that interval. These values can be the endpoints.

continuousopen interval

no maxno min

discontinuousclosed interval

has maxno min

Let f be continuous on [a,b] and differentiable on (a,b). If f(a)=f(b) then there is at least one number c such that f’(c)=0.

CritiCaL number

extreme VaLue theorem

roLLe’s theorem

12

Assume that f(x) is differentiable and let c be a critical number of f(x)

If f’(x) changes from + to - at c THEN f(c) is a real maxIf f’(x) changes from - to + at c THEN f(c) is a real min

Assume that f”(x) exists for all in (a,b)

If f”(x) > 0 THEN the graph is concave upIf f”(x) < 0 THEN the graph is concave down

If f”(c)=0 and f”(x) changes sign at x=c, then there is an inflection point at x=c

1) y=x5tan x

2) 2-x3x+1

Find y’.

3) 3x3+4y2

first DeriVatiVe test

test for ConCaVity

test for infLeCtion Point

PraCtiCe ProbLems

13

y =

y =

antiderivatives

14

The antiderivative or indefinite integral of a function f(x) is a function F(x) whose derivative is f(x). Since the derivative of a constant equals zero, the antiderivative of f(x) is not unique; that is, if F(x) is an integral of f(x), then so is F(x) + C, where C is any constant. The arbitrary constant C is called the constant of integration.

f(x) dx=F(x) + C

integrand

variable of integration

constant of integration

kf(x) dx=k f(x) dx (k=0)

[f(x)+g(x)] dx = f(x) dx + g(x) dx

=ln u +Cduu

cos u du = sin u+C

sin u du = -cos u+C

tan u du = ln sec u +C

cot u du = ln sin u +C

sec2 u du = tan u +C

csc2 u du = -cot u+C

sec u tan u du = sec u +C

csc u cot u du = -csc u +C

sec u du = ln sec u + tan u +C

csc u du = ln csc u + cot u +C

eu du = eu +C

au du = +C (a>0, a=1)au

ln a

What is a antiDeriVatiVe?

basiC antiDeriVatiVe formuLas

un du = +C (n = -1)un+1

n+1

15

_ _

0dx = C

kdx = kx+C

kf(x)dx = k f(x)dx

f(x) g(x)dx = f(x)dx g(x)dx+_+_

xndx = +C= xn+1+Cxn+1

n+11

n+1

A) 3 x dx= x1/3dx = x4/3+C = x4/3+C14/3

34

B) 1x2 dx = x-2dx = x-1+C = +C-1

1-1x

1. Look for a piece of the function whose derivative is also in the function. If you’re not sure what to use, try the denominator or something being rasied to a power in the function.

2. Set u equal to that piece of the funciton and take the derivative with respect to nothing.

3. Use your u and du expressions to replace parts of the original integral, and your new integral will be much easier to solve.

A) 2x(x2+1)4dx = u4du= +C = (x2+1)5+Ctu5

515

u = x2+1du = 2xdx

B) 3x2 x3+1 dx= u1/2du = u3/2+C = (x3+1)3/2+C23

23

inteGration ruLes

ExamplEs

u-substitution

ExamplEs

16u = x3+1du = 3x2dx

xndx = +C= xn+1+C

If a function f is continuous on the closed interval [a,b] and F is an antiderivative of f on the interval [a,b] then

To find exact areas under curves, use definite integrals. The area beneath x2+1 on the interval [0,3] is equal to dx.(x2+1)

3

0

f(x)dx = 0a

a

f(x)dx = a

bf(x)dx

b

a

_

f(x)dx = b

af(x)dx

c

af(x)dx a<c<b

b

c+

kf(x)dx = kb

af(x)dx

b

a

[f(x) g(x)]dx = b

af(x)dx

b

ag(x)dx

b

a+_ +_

f(x)dx = F(b)-F(a)b

a

f(x)dx = F(x)b

a

b

a

1) (3x2-2x+3)dx

Answer these questions.

3) 4-2t dt

Definite inteGraLs

ProPerties of Definite inteGraLs

funDamentaL theorem of CaLCuLus

PraCtiCe ProbLems

17

2) (2-3x)5dx

answers

to

probLems

18

0-114

1)2)3)

By the Product Rule:y’=x5(tan x)’+(x5)’(tan x)y’=x5sec2 x+5x4tan x

By the Quotient Rule:

y’=

y’= -

(3x+1)(-1)-(2-x)(3)(3x+1)2

7(3x+1)2

9x2+8y dydx

1)

2)

3)

x3-x2+3+C1)(2-3x)6+C-1

182)

u=4-2tdu=-2

4-2t (-2dt) = +C-12

-12

(4-2t)3/2

3/23)

Limits & Continuity

DeriVatiVes

antiDeriVatiVes

19

What is a limit? PaGe 3

What is a derivative? PaGe 9

What is the first derivative test? PaGe 13

What is an antiderivative? PaGe 15

What are definite integrals? PaGe 17