2.1 The Tangent and Velocity Problems Ex: When you jump off a swing, where do you go? Ex: Can you approximate this line with another nearby? How would you get a better approximation?

lF±¥# is.t€*t,

¥¥•÷rhe*¥

v.tt

Ex: A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after 5 minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. t (minutes) 36 38 40 42 44 Heartbeats 2530 2661 2806 2948 3080 The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient’s heart rate after 42 minutes using the secant line between the points with the given values of t. a) t = 36 to t = 42 b) t = 38 to t = 42 c) t = 40 to t = 42 d) t = 42 to t = 44 What are your conclusions? Notice in this example that we calculated slopes, which gave us ∆"∆# =

= So, the slope told us about the rate of change of the graph.

m - any ¥530#f42-36 secant= 4161 = 69.5 beat per

minute

m= FIT = 29¥66'= ¥7 = 71.75k¥ min .

m= an¥= 29*06=142-2--71 bpm

m=an¥= 30¥48 = l3÷= 66 bpm

Change in total heartbeats Away )

change in timeHeart rate

Units: Position: Velocity: Acceleration: Average Velocity: Instantaneous Velocity: Ex: If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height in meters t seconds later is given by y = 10 t – 1.86 t2. a) Find the average velocity over the given time intervals: i) [1, 2] ii) [1, 1.5] iii) [1, 1.1] iv) [1, 1.01] v) [1, 1.001] b) Estimate the instantaneous velocity when t = 1.

m = metersft= feetcm= centimeters

ftp.ft/m..n.kmls ¥m/s2= %

, 9.8%2ftlsr

Velocity calculated between two points .

The velocity at a given point

F. s Img .siMd

1 : y= 10 .CH1.8645=10 -1.86=8.14r 2 :

y= 10 (2) - 1. 8644.20-7*44=12.56± tf¥= ftp.4 = 4.42%

. s.

e ~ aatz.ly?t5jI=2o6#=5.3sm1s§ u±÷I÷iE÷EE:O:*o 3 # = - = to} at 1.01-1)

8.14628-8.140.00628=6.28 %£ 4 AI=

foot iooAt

= 6.28 Mlsor = 6.3 or = 6.29

Given: 1) Any letter can change into an A. 2) Two Cs in a row can turn into a B. 3) Any word can be doubled. Ex: C ACAC Ex: CAC CAAAC Ex: C BACABC

Any word can be # 4⇒ octupled

CA # I C C # 3

Fatma##} ,| £cc#¥3 | A C # 1

ACAC # 1 A CAC # 3o o

# QED

CACIAC # 3

CABAC # 2

CAAAC # 1

0

CC # 3c C c C # 3

C CCC Cccc # 3

CCACACCC # 1

# 2

2.2 The Limit of a Function Ex: Graph this function: & ' = '(&)*' ≠ 2

1) What is f(1)? 2) What is f(3)? 3) What is f(2)? 4) What would you guess f(2) should be? 5) Fill in this chart:

x f(x) x f(x) 1 3

1.5 2.5 1.6 2.4 1.7 2.3 1.8 2.2 1.9 2.1 1.95 2.05 1.999 2.001 1.9999 2.0001

"1

YH×< •

✓

1

9

Undefined

4

1 92.25 6.252.56 5.762.89 5.293.24 4.84

3.61 4.413.8025 4.20253.996001 4.004001

3.99960001 4.00040001

. .

The left side of the table tells us that lim0→(2

&(') = 4 which is read as “The left-hand limit of f(x) as x approaches 2 is equal to 4” or “The limit of f(x) as x approaches 2 from the left is 4” The right side of the table tells us that lim0→(5

&(') = 4 which is read as “The right-hand limit of f(x) as x approaches 2 is equal to 4” or “The limit of f(x) as x approaches 2 from the right is 4” Note that x→ 67means “from the left” and not necessarily negative numbers. The same is true of x→ 68 In this case, the left-hand limit and right-hand limit are equal, so we can say that lim0→(

&(') = 4

In general, 9:;<→=

>(<) = L if and only if 9:;<→=2

>(<) = L and 9:;<→=5

>(<) = L

Example 7, page 92:

a) lim

0→(2? ' = d) lim

0→@2? ' =

b) lim

0→(5? ' = e) lim

0→@5? ' =

c) lim

0→(? ' = f) lim

0→@? ' =

time limit.

The limit as × approaches 2 from the left

(2,3 )

( 5,2 )

G. , )( 5,1)

# ¥

3 2

I 2

Does Not z g( 5) =/Exist

DNE

Infinite Limits Ex: Find lim

0→AB0C if it exists.

Approach 1: Table of values (Why can we use “±”?)

x 1'(

±1 ±0.5 ±0.2 ±0.1 ±0.01 ±0.001

Approach 2: Sketch the graph Hmmm…it looks like f(x) keeps getting bigger and bigger as x approaches 0 from the left and the right. In this case we say:

lim0→A

1'( = ∞

Definition: Let f be a function defined on both sides of a, except possibly at a itself. Then lim

0→J& ' = ∞

means that the values of f(x) can be made arbitrarily large (as large as we please) by taking x sufficiently close to a, but not equal to a.

Note: Some texts phrase this as “the limit does not exist.” This can cause confusion, since that also refers to the case when the left and right hand limits do not agree.

14

25

100

10,0001,000,000

M¥.

Nole : Some books say thismeans the limit DNE

Ex: Find lim0→K2

007K , lim0→K5

007K and lim

0→@0

07K

t.sn#=sIy=F=5oxlFxEEyE.8.gn!

Try �1� Tables or @ Graphs Y=¥×=4 Not to scale

y

%¥÷e÷*×t.gg#=.oolinx.y=ao×→4t

s¥¥¥÷⇒¥+9

335¥ ,=Io¥= -7

3.9 ?¥,=to÷= -39

3.99|÷If=?oIt=-399

±-

=

.

2.3 Calculating Limits Using the Limit Laws

Limit Laws: Suppose that c is a constant and the limits lim

0→J&(') and lim

0→J?(')

exist. Then 1) lim

0→J[ & ' + ?(')] = lim

0→J&(') + lim

0→J?(')

2) lim

0→J[ &(') − ?(')] = lim

0→J&(') - lim

0→J?(')

3) lim

0→J[P & ' ] = P lim

0→J&(')

4) lim

0→J[ & ' ∙ ?(')] = lim

0→J&(') ∙ lim

0→J?(')

5) lim

0→J[ R 0S(0)] =

TUVW→XR(0)TUVW→XS(0)

if lim0→J

?(') ≠ 0

Ex: Use Law #4 to show that:

lim0→J

[ & ' ]n = [lim0→J

& ' ]n (6 - Power Law)

Slept : Prove for n=l

xjmaffh'D 't[ lxigafh'D'

lxisma GAD '=ti→mafG)'E¥po0fnts

True

fnjnafcx) = [lizfGD"

" ← be ,

Step 2 : Assume for n=K Assume

lxjna [ FHDK - [ fszfcx) ]k True

step 3. Prove for n=ktl

←

tiga [fC×Dk" I[ lxignafcx)]k+

1

Step I : Prove for n= 0 or n= I

Step 2 : Assume for n= K

Step 3 : Prove for n= Kt I

ftp.#xDxwcxDff=xbimaffADklismafCx)Hgggit-.=fxiyafCxDkt

'

Additional Laws:

7) lim

0→J' = c

8) lim

0→J( = a

9) lim

0→J(Y = +Y

10) lim

0→J([ = +[ (As long as +[ is defined.)

11) lim0→J

-(()[ = lim0→J

-(()[ (As long as lim0→J

-(()[ is defined.)

Ex: Evaluate these limits, justifying each step. 1) lim

0→\(2(( − 3( + 4)

2) lim

0→\(0C7\08K(07\

=lxig2x2×ljg3xtdjms4 Laws

1+2=21;gxt×3jgx+×hjm4 law

3=2*3×3'

-3*3×+1*4<w6

=2[ 3 ]t3( 3) + fight Low 8

=2[ 3) 2- 3( 3) +4 Law ?

= 18-9+4=13

nineIxlijafcxhfca)

atxtgma .f(D= f (a)

lxignafcx) # FG)xk→maf(×)DNEx==÷sina.tk#xgma+Fhs

lim0→\

(2)( − 3) + 4): 1) lim

0→\(2)( − 3) + 4) = lim

0→\(2)( − 3)) + lim

0→\4 Law #1

2) lim

0→\(2)( − 3)) + lim

0→\4 = lim

0→\2)( − lim

0→\(3)) + lim

0→\4 Law #2

3) lim

0→\2)( − lim

0→\(3)) + lim

0→\4 = lim

0→\2)( − lim

0→\3 ∗ lim

0→\) + lim

0→\4 Law #4

4) lim

0→\2)( − lim

0→\3 ∗ lim

0→\) + lim

0→\4= 2lim

0→\)( − lim

0→\3 ∗ lim

0→\) + lim

0→\4 Law #3

5) 2lim

0→\)( − lim

0→\3 ∗ lim

0→\) + lim

0→\4 = 2[lim

0→\)]( − lim

0→\3 ∗ lim

0→\) + lim

0→\4 Law #6

6) 2[lim

0→\)]( − lim

0→\3 ∗ lim

0→\) + lim

0→\4 = 2(3)2 – 3*3 + 4 = 13 Law #7, Law #8, Arithmetic

lim0→\

2)( − 3) + 42) − 3

1) lim

0→\(0C7\08K(07\ =

TUVW→a((0C7\08K)

TUVW→a((07\) Law #5, plus lim

0→\(2) − 3) ≠ 0

2)

TUVW→a((0C7\08K)

TUVW→a((07\) = B\

TUVW→a((07\) (Problem above)

3) B\

TUVW→a((07\) = B\

TUVW→a(07TUVW→a\ Law #2

4) B\

TUVW→a(07TUVW→a\= B\

(TUVW→a 07TUVW→a\ Law #3

5) B\

(TUVW→a 07TUVW→a\ = B\

(∗\7\ = B\\ Law #7, Law #8, Arithmetic

Ex: Find lim0→7(

0C7K08(

Ex: lim

b→A\8b C7c

b Ex: lim

0→B) + 1EF) ≠ 1GEF) = 1

One-Sided Analysis Remember that: IJK

L→MN(L) = L if and only if IJK

L→M2N(L) = L and IJK

L→M5N(L) = L

Ex: Show that lim

0→A) =0

step I : Plugin 8-'

n'

die "¥x¥t - tiazaas= - 2-2=-4

\ I 1 /

8- ooo

= thing9+6h+h÷

=tin 6h + hi

h→o T

= his hC6t¥= thing ( 6th ) = 6

*a¥eei÷iY

1×1 = {× if xso

- × if × < O

lim 1×1 :

⇐xia

' ''

Ktisnofxtoor = - 1 lixgfx = - 0

⇐ +

w±x¥¥¥Io:o)jygolxl = 0

Ex: Prove that lim0→A

00 does not exist.

Ex: If f(x) = 2 − )+,) < 2

)+,) ≥ 2 , determine whether lim0→(

,()) exists. Greatest Integer Function [[x]] = The largest integer that is less than or equal to x. Ex: [[5]] = 5 [[2]] = 3 [[ 3]] = ?

tin .

¥ itsmo¥

=1 in C-D= - 1

× → o- i

tsmot ¥ = Has ¥ = txisno ' = 1

figs f G) = xhjz .

a

= tax× → 2-

= F = for = 0

time fkk dig × = 2 cnn.fmSame

fig ffx ) D N E

z -

o_0Eyes%ni⇒I¥¥g[ [ - e ] ] =-3

Theorem: If f(x) < g(x) when x is near a (except possibly at a) and the limits of f and g both exist as x approaches a, then lim

0→J,()) < lim

0→J7())

Squeeze Theorem: If f(x) < g(x) < h(x) when x is near a (except possibly at a) and lim

0→J,()) = lim

0→Jℎ()) = L

then lim

0→J7()) = L

Ex: Show that lim

0→A)(9+: B

0 = 0.

-x2 <)(9+: B0< x2

Summary of Techniques So Far: <=>?→@

A(?) 1) When in doubt, look at the graph. 2) If the function is defined at a, try substituting a for x. 3) If the function is not defined at a, try expanding and/or factoring to cancel out

problem factors. 4) If you suspect that the limit might be infinite, look at a sign analysis/sign chart. 5) If you suspect that the limit might not exist, consider a one-sided analysis. 6) If you can find g(x) > f(x) > h(x), use the Squeeze Theorem.

2.4 The Precise Definition of a Limit (Delta-Epsilon Proofs) What does “x gets really close to a” actually mean? The distance between x and a gets smaller and smaller. How do we express the distance between x and a? We use the Greek letter B(CDEFG) to express an upper bound for this distance:

| x – a | < B What does “f(x) approaches a limit” actually mean? The distance between f(x) and the limit, L, gets smaller and smaller. How do we express the distance between f(x) and L? We use the Greek letter H (epsilon) to express an upper bound for this distance:

| f(x) – I| < H

In fact, if the limit actually exists, we should be able to get arbitrarily close to L as x gets closer to a. So we start with a value for H, and then solve to find a B that works. Translation: Step 1: Pick any value for H that you’d like, or use one that someone else gives you. (It has to be positive, and usually it’s between 0 and 1 – often very close to 0.) Step 2: Set up the inequality L – H < f(x) < L + H Step 3: Solve to find the interval of x values that will satisfy the inequality. Step 4: Use this interval to come up with an appropriate B.

nm

÷ftp.ex

Ex: For f(x) = x + 2, a = 1, and L = 3 find a B that corresponds to H = 0.01 Now it’s possible that H = 0.01 was a special case, so what we really want to do is to look at all possible values of H, not just specific ones. Ex: f(x) = x + 2, a = 1, and L = 3, find a B that corresponds to all values of H Ex: Prove that lim

0→(3) = 6 (for all values of H)

type py.

value xa,

o

y :L ,E

×<=?x,= ?netted"

I%ff.is#IoIE43sa.iH=aomotCxz ,

-14=(0*2.99)limfcx)=L ×

3x→a 0.9+94<1.01

lxjm,fC×)= 3 p8=0.01×-5xts

we want to find J sit. toragivn value

of E,

if /× - akd

then lfcx ) . LKE

1×+2 - 3/< E

|wengeedtfndo

1×-11 - E 1×-11<8gets us there- { < xt<E←nyis5k¥

⇒ D= E

lygafcxkl Step 2 : Write out

We need to find or /ka( < S

st 1×-21<0if .lx . akd

step } :D#hthen IFCH . LKE

Make theresultsStep 1 : Start with

ofstpl look likelfcx ) - L|< E STPZ

13×-614 13×-61<231×-4<2 Let

1×-21<5 ← o=§.

1×-51 §I

iiIs÷y"→k¥±ah

Infinite Limits Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then lim0→J

' ( = ∞ means that for every positive number M there is a positive number , such that if 0 < | x – a | < , then f(x) > M

(Note that an infinite limit technically means the limit doesn’t exist, just in a particular way.)

Ex: Prove that lim

0→AB(0C = ∞

We want to find or st .

( P" 6)

if 0 < 1×-4<5then f ( × ) > M

Step I : We want f ( × ) > M

¥2 > M

Step 2 : We need to find of

st .

0<1×-0 / < or

0 < 1×1<5

Step 3 : Dea#

2¥ > M

¥s¥mai*Ij#¥aWe shouldchoose

f- Ff=

FmTm

1/2<4

If ×

>0 | If

XEOO£×<2-2 <* 0

1×1<2

2.5 Continuity

A function f is continuous at a number a if lim

0→J' ( = '(+)

This means three things: 1) f(a) is defined 2) lim

0→J' ( exists (and is finite)

3) lim

0→J' ( = '(+)

Ex: Draw examples of ways in which a function could be discontinuous. Removable discontinuity: Jump discontinuity: Continuous from the left: lim

0→J2' ( = '(+)

Continuous from the right: lim

0→J5' ( = '(+)

÷÷o÷

-

Ex: Where are each of the following functions discontinuous? a) f(x) = 0

C707(07(

b) f(x) = B0C 4'( ≠ 014'( = 0

c) f(x) = 0C707(07( 4'( ≠ 214'( = 2

d) f(x) = [[x]]

€fcxj.tk#ylYI=x+lxt2X=Oyfdiscb@x=Oc1f=D:R

t.EE?s#I?IE***y

p 105

t.I.it#MxDisaaQz,or Frasers

Theorem 7 - The following are continuous at every value in their domains: 1) Polynomials 2) Rational Functions 3) Root Functions 4) Trig Functions 5) Inverse Trig Functions 6) Exponential Functions 7) Logarithmic Functions Ex: Evaluate lim

0→rsUt 0

(8uvs 0 Steps: A) Is the numerator continuous? B) Is the denominator continuous? C) Are there any points where the rational expression isn’t defined? => If the function is continuous everywhere, then lim

0→r'(() = f(@)

Theorem 8 – If f is continuous at b and lim

0→JA ( = B, then

lim0→J

'(A ( ) = '(B), which means lim0→J

'(A ( ) = '(lim0→J

A ( ) (See Example 8, Page 124)

: - SINX

Yes

Y= Ztcosx;at is defined for b to

Ztcosx ¥0

findF¥s×=In÷←=E

,-÷.

Ex: Show that there is a root of the equation 4x3 – 6x2 + 3x – 2 = 0 between 1 and 2.

P ( x ) = 4×3 . 6×2+3×-2polynomial

P ( D = 4 ( 1)3 - 642+3 ( i) -2

= 4 - 6+3-2=-1P (2) = 4 (2) 3- 6( 2) 2+3 (2) -2

EYE: EIY III. a

lI÷I.IT#tEEiEo.

the x - axis

between * I IX= 2

Common Mistakes

1) lim0→(

0C7@08x07( =

AA DNE

Hint: Always check to see if something will cancel 2) lim

0→A(0C =

(A DNE

Hint: If nothing will cancel, check the graph

3) lim0→(

0C7@08x07( =

07( (07\)07( = 0 − 3 = 2-3 = -1

Hint: Know how formal your professor is

bar "¥÷ 's.ae#F==xljn(x-3)=2

-3=-1→

adf.lt#ksmoia=oliner liner

a ×¥z÷tia"→¥⇒=I→z(×-3 )

= 2-3=-1 Not necessary

#to

write

I¥E÷÷EI¥÷I×¥¥n¥Y¥÷O×→2

ay÷i¥*⇐y¥i÷i¥¥÷-3

for × # 2

<j÷*'s

t

2.6 Limits at Infinity – Horizontal Asymptotes Look at the graph of f(x) = B0 As x -> +∞, f(x) -> As x -> −∞, f(x) -> As x -> 0+, f(x) -> As x -> 0- f(x) -> The line y = 0 is called a horizontal asymptote of f(x), since lim0→x

+ , = 0 Note: lim

0→x+ , = / can be read as:

1) The limit of f(x), as x approaches infinity, is L 2) The limit of f(x), as x becomes infinite, is L 3) The limit of f(x), as x increases without bound, is L

Theorem 5 If r > 0 is a rational number, then lim0→x

1,y = 0

If r > 0 is a rational number such that ,y is defined for all x, then lim0→7x

1,y = 0

g.to#xE*ror*o:

- -

Ex: Evaluate lim0→x

\0C707(@0C8K08B (Using Theorem 5 above)

(Note: Limit Laws only hold for finite limits, not infinite limits, so use the above technique to rewrite the following function.) Ex: Evaluate lim0→x

(0C8B\07@ (Note: ,( = ,

= * a ¥s¥¥¥¥t¥

= * . F÷¥¥*

t.IT#o=t5pppoohgggpgftM@⇒

sF÷IFi¥a¥F¥I=¥

Ex: Evaluate lim0→x

( ,( + 1 − ,) Hint: Use the conjugate. Ex: Evaluate lim

0→x(,( − ,) Hint: Look at this as a product.

Ex: Evaluate lim

0→A5=>?@=A B

0 Hint: Let t = 1/x, and then look at the graph of arctan

( ttf . x ) txftx ) linked?A5link fat)=×→axFtx

Cat B) CA - B) = AZB'

= him ×2+l#×→• Kat X

=ti→a×¥x .

tismalx'

. xktjzxcx . D

€

= A

lxignoarctantx = lim arctanttsoo

LettxtAs x→ot

,t=tx→ a

=lim want =/.bz?I2t→ -

From the

graph

Formal Limits –

Let f be a function defined on some interval (a, ∞). Then lim0→x

+ , = ∞ means that for every positive number M, there is a corresponding positive number N such that if x > N then f(x) > M

Ex: Prove that lim0→x

C0 = ∞ Step I : Given M

,we want to find

NS.t .

if × > N =) ex > M

We want

ny.ie"

ex > M r

.

...>

/ne× > 1mm t.ie#is:IX > lnm y.mx .

Let N=1nM . isincreased'

soit

preserves

then,

if × > N=hMorder

=) ex 's M

Summary - Visual Representations of Limits 1) Finite limits as x approaches a

lim0→J

' ( = * means that for every + > 0,

there is a number , > 0 such that

if ( − 1 < , then '(() − * < +

(Page 111)

(Page 112)

lxljmafG) =L

:

:

:

•

2) Infinite limits as x approaches a

lim0→J

' ( = ∞ means that for every 5 > 0,

there is a number , > 0 such that

if ( − 1 < , then ' ( > 5

lim0→J

' ( = −∞ means that for every 6 < 0,

there is a number , > 0 such that

if ( − 1 < , then ' ( < 6

(Both images, Page 116)

lxijnaF G) = a

lijnafcxt . a

3) Limits as x approaches infinity/gets large without bound:

lim0→y

' ( = * means that for every + > 0,

there is a number N such that

if ( > 6then ' ( − * < +

lim0→7y

' ( = * means that for every + > 0,

there is a number N such that

if ( < 6then ' ( − * < +

(Both images, Page 138)

lxisnoof CD =L

txignafcxk L

lim0→y

' ( = ∞ means that for every 5 > 0,

there is a number N > 0 such that

if ( > 6then ' ( > 5

(Page 140) djgzfcx)= a

2.7 Derivatives and Rates of Change Let’s look at secant lines again: For a given curve y = f(x), we’re interested in what the tangent line at P (a, f(a) ) looks like. So we look at another point, Q (x, f(x) ), to the right of P, and investigate what the secant line through P and Q looks like:

Since we now understand a little about limits, we can look at what happens as we move Q closer and closer to P by letting x approach a. The slope of PQ = ~;ÄÅ~ÇY =

∆É∆0 =

R 0 7R(J)07J

so, as we nudge Q closer to P, we get

.

A = delta

ay m=AY*ax

m = lim0→J

R 0 7R(J)07J

This is the slope of the tangent line at P. (Provided the limit exists.) Ex: Find an equation for the tangent line to the parabola y = x2 at the point P(1, 1). If we look at the original idea of points P(a, f(a) ) and Q(x, f(x) ), but, instead, consider the x values as x and x + h, the new graph looks like this:

4¥ limipgyttsmope

m= tinyftp.Ia#x2.fLa=1=tisx-=tig'¥ '

=tig*¥¥=lxigCx+D=2m=2 ,

( 1,1 ) y - 1=2 ( x - 1) ⇐ tenant 4=2×-2+1

4=2×-1

m =3 ( l,

2)

Point - slope :

y- y ,

= m ( × - ×, )

viable t#y - 2 = 3 ( × - l )

y - 2 = 3 × - 3+ 2

+ z

y=#i< Slope - Intercept

I to,

- D

Ex: (#14 in the book) If a rock is thrown upward on the planet Mars with a velocity of 10m/s, its height (in meters) after t seconds is given by H = 10t -1.86 t2. a) Find the velocity of the rock after one second. b) Find the velocity of the rock when t = a. c) When will the rock hit the surface? d) With what velocity will the rock hit the surface?

t= OsH=

Ont.ISH = 10-1.86=8.14 m

Average Velocity = st# = 8.14 %

t=O H= 0

t=a H=lOa- 1.86 a'

Average Velocity =( 10 a -1.86 a

' ) - 0

To= 109 - 1.86 al

a-=

10 - 1.86 a Mls

§jh&EkIgs'i¥I:⇐o±A¥I#→

tdo-1.864=0toor

10-1.86+-05t = # s

It should be -10 m/s

Velocity = 10 - 3.72T

= 10 - 3.72 (F) = 10 - 20=-10

To get a good approximationfor the instantaneous Velocity ,

pick pts close to t=a is calc.

the average velocity.

t- a HH .

- lot -1.86T'

t= ath

H (a) = 10 a -1.86 a2

H ( at D= 10 ( ath ) - 1.86 ( ath )'

= 10 a + Koh - 1.86 ( dtzahth ' )

=

Hcath ) - H (a)Average -

Velocity ( ath ) - a

= www.t#nxa=yf*aaY

=10h-3.72aht.SI

hI

=X(10 -3.72A - 1.86k )

#=

-10-3.74-1.864<7A. velocitybetween a and

ath

Instantaneous I,

velocity = njmo(Average Velocity )

=ntgm (10-3.74-1.864)

= 10 -3.72A

2.8 The Derivative as a Function

!′($) = limb→A

! $ + ℎ − !($)ℎ

Ex: If f(x) = 2x2 – 4x + 3, find f'(x). Notation: f'(x) = y'=ÖÉÖ0 = ÖRÖ0 = ÖÖ0 f(x) = D f(x) = Dx f(x) = first derivative f''(x) = Ö

CÉÖ0C =

ÖÖ0 f'(x) = second derivative

*a×=

riserun

ftp.nliggfcxth#H=LignoEAthF4Cx+h)+D- [2×2-4×+3]

=

=L;no#4×htzh¥*4_I#*8=

tin 4×42*44h→ O

,

= Liz H4xt# )

= nlism (4×+24-4)=4×+0 - 4= 4×-4

← n¥x

¥⇒.

A function f is differentiable at a if f'(a) exists. It is differentiable on an open interval (a, b) or (a, ∞) or (-∞, a) or (-∞,∞) if it is differentiable at every number in the interval. Ex: Where is f(x) = |x| differentiable? Some ways a function can fail to be differentiable: Corner/kink: Discontinuous: Vertical Tangent Line (with or without a corner):

.IE#II*,

Yetn

:&

Ex: If f(x) = 4x3 – 2x2 + 3x – 12, find f'(x), f''(x), f'''(x) and f(4)(x)

Ffx )= him fcxth ) -

f#h→O

= ,.in#xthF2lythIDDF_4x22xHx.iDh→O

-

=L;z[4Cx3+3×4+3 'xh2thY

-2×2-4×4-2443×+4-1%3

-

=L;g4×412×4+12 xht 4h

'

-2×2-4×1-2443×+34-12

-4×12×33×+12

== lim

12×2hH2xh44h3-4xhth2+h→0, h

qfijno X(Rx4Rxh+4h2.4×-24+3)

#= fig

(144121+412.4×-21+3)

=

12×2-4×+3

..

f 4×1=12×2-4×+3

f"

( D= Lsgfkthhtftxj=hipo[24+42.4*4+3=[12×2-4×+3]

.

- nlingo #24×h+kE#4ht3x#xX-

=tin

24×4+125. 4h

hso == high (24×+124-4)

== Lingo (24×+124-4)=24×-4

f " '

( × ) = lim f"C×th="Cx)hso h

qnhjmo[246+4.4]-[24×-44

24×+24 h - 4-24×+4= Ko -

= him 2¥ = Liz 24=24

f " )(×)= 1

;mf"Gth#(jhe- tie2k¥ Koen

= Lingo 0=0

f ( × ) = 4×3 - 2×2+3×-12

f ' G) = 12×2-4×+3

f "

( x ) = 24×-4f

"

G) = 24 ← 24×0

f ' " ( x ) =o

.

Pascal 's A

111( xth ) '=×+h

1,,2,)

(×+h)÷x72xh+h'

I 3 3 I ( x+h)?xI3x'ht3xh2+h3\ , - in ,

I 4 6 Y I

(xthjhx 't 4×4+6×42+ 4×h3+h4