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I (a) y = 4x + 9x outside (^) -> In inside ->^ 4x+ 922 (b)y

an(n2+ 1) outside ->^ tanz inside ->n+ 1 (2) y = secm outside (^) - i I inside ->^ sec u (d) y^

(

ex)

outside ->^ i inside

• (^) He

186 (see f(x) =^ ((n

• v2)(x+ 2) -^ (n^ + 1) (x

A(x +^ 1) #y =^3 (Ay

tH. An+1+y = 3Ay

• (^) 15x + A. Ar-3Ay+

A =

13x-y-1. A(x

• zy + 1) =15x - y- 1. A

· Find an (^) equation for the^ tangent lineat (^1) , ) y = f'(a)(n

• a) + f(a) n

y

3ny y- I

• = 3 3 y = 3xy

x x- 3 y = Sy

• 3n - 3 iyz = y(y

n) n2 + yz = y

• x ya

y = x - x y(y

= x(1 - x)

. f(x) g()

· Me M ↓o (^) Jr. W

i ginegist

I ↓ I IAL

-^ I & I I 'I' : I 1 I 1 I

T EST

REVIEW SET^ OF^ PROBLEMS^ ·

Absolut maximum (maximo^ absoluto) ·Extreme values (Absolut (^) minimum (^) (minima relatio

81. Sketch^ the^ graph of^ a^ continuous^ function (0,4)^ having a local minimumbutno absoluteminimum. I

(^3) - (^2) - (^1) -

I 1 I - x

35. Two^ days (^) after he^ boughta^ speedometer for hisbicycle,

Lance broughtitback^ to^ the^ yellow^ Jersey Biker^ shop. "There^ isa

problem withthisspeedometer,"fance^ complained to^ the^ clerk. "Yesterday Icycled the^ 22-mileRogadzo Road^ Trailin^78 minutes. and notonce^ didthe^ speedometer read^ above^15 milesper hour!" "Yeah?"responded the^ clock.^ "That's^ the^ problem?"How^ didLance use the^ Mean^ Value^ Theorem to^ explainhiscomplaint? s =^22 miles^ #15miles/h t = 78 minutes v =^ S^ -^ -> E v = Mut: If I^ iscontinuous on^ Ia, b) and (^) differentiable on (^) (a, b), then^ there^ existsatleastone value a in (^) (a, (^) b) such that f(x) = 1 = f(b) - f(a) = f()(b

a)

(^15) miles/per how

SECTION 4.

· 21. The^

growth of a^ sunflower^ during the first (^100) days after sprouting^ is modeled^ well^ by the log istic^ case^ y:h(t).^ Estimate^ the growth

mate at

the (^) point (^) of inflection and (^) explainits significance. Then make a rough sketch (^) of the^ just and sewerd h dairatives of

h

D (^) h(t) (^250) - Near the^ point of inflection,^ the (^) curre (^200) - 140,150) is roughly a straight

line going

(^130) -------.(((r) =^0 through (33,200) and (^) (33,100) (^100) - "

pointof inflection^ to^ pate^ of change=

e (^30) - I (^) and so (^) ob Ibot 20

& D Rate of change:

• ↑ L a

max

• I (^) I I^118 hi(40) en #no so on^ is ty=ht) S "E I 11 11 t

⑳

39. Wateris (^) pumped into (^) a (^) sphere (^) of radius R ata variablerate^ in such (^) a (^) way thatthe water level rises ata constant rate^ (Figure 20). LetVIt) be the volume (^) of waterinthe^ tank^ attimet. Sketch^ the graph (approximately,^ but^ withthe correct (^) concarity). Where does the (^) point (^) of

inflectionoccur?

V

·^ inflection

t SECTION 4.3^ ~ In (^) Exercises15-36, (^) find the (^) transition (^) points, intervals (^) of increase/ decrease, (^) concarity, and^ asymptotic behavior.^ Then^ sketch^ the^ graph, withthis informationindicated. 27.y =^ x^ -^ 4F^ ·^ Ata = 0 the (^) derivativedoes not exist y

=

F D(9)^ = [1, + 3) y

2 =^0 = ) = (^1) =) (^) F(2)-=) m =4. 1 4

• (^) 3. I is^ increasing · a at^ (4, +0)^ since^ is positive. y

• t ·

I is^ decreasingat^ 21,4)^ Since^ y'is negative.

I

W &

D(y) = (0, + 0) to^ him^ f(n) =^ - > Asintotavertical x-^ -^ D

• (^312) y = 1

2a(y" = 0 = 31 - x =^0 = 11 - 5 =

7 (0,

conce up = 1 - 2 = 0 = 31 - F = E = E = 0 = r= 0 = (n

......

im f(n) = - x Asutotavertical

n - -^ - tim (^) f(x) =^0 Asintota horizontal n -^ i

SECTION 4. 1

11. a^ = 3, Ax^ =0.

f(x)

I ((n) =^ ( -( +x2) 4 19:f'(a)^

An (

+ m)

(v +a) eri-Tasz en- ?(t) e -1 -^ 12 (a) (1+2) (^) nFr

By = - 5.0.3=-0,03125.cal-10/

• 0,0313 me^ (aprox.) A = 10,625.0, = - 3,
• 3,13 - (- 3,125) - 0,03125- (-0,0313)^

= error.

=- 3,13 +^ 3,125^ =^ -0,

e If^

y'(a) De A t^ h = 20km, P (^) =5.3 (^) kilopascals. ·- f) I Estimate^ P^ at^ altitude^ 20.3^ km.

2. EllimateAP^ at^ h^ = 20 when (^) Ah (^) = 0. P = 3.3 + 0.5)-0.87) =3.065 (^) kilopascals. 14 = (^1) 0,87).0,

we

a

20.3) th

47. (^) =25cm.
    • Estimate the (^) maximum error (^) inthe (^) volume and surface are i m is^ accurate to^ within0.5cm. AV = 4 π^ (25)(0.5) AS = 8iT (^) (25) (0.5)

v =

2 πr

3V

24(25) 3D R = 10. 3 = 4 πr2 (^) S = 4 π(25) If R^ =^25 then, xf^ = y'(a)Bm

AV =^ v'(25)^ Dr

E As =^ S'(23)Ar danuts