ANALYTICAL APPLICATIONS OF DIFFERENTIATION NOTES...EX#2: Let/ be the function given by f(x) = x 2 +...

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UNIT 5:ANALYTICAL

APPLICATIONS OF DIFFERENTIATION

NOTES

𝑓𝑓′ 𝑐𝑐 =𝑓𝑓 𝑏𝑏 − 𝑓𝑓(𝑎𝑎)

𝑏𝑏 − 𝑎𝑎

a c b

f(a)

f(b)f(c)

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EX#2: Let/ be the function given by f(x) = x2 + 2x - 3. Find the number c that satisfies the conclusion of Rolle's Theorem for f on [-3, 1].

\) f/(x)=�x+J J(x+l) =O

JJ f(-3) = f(1) =O

X = - I

J) + is Contf nuou.s

+ d;fferentia.b)e(pol�Y\omioJ)

�) f (I) - f ( - 3) - J X + � \ -(-.3) -

when x = - I

5) : .. C -=- - IEX #3: A sports car speedometer registers 50 miles per hour as it

passes a patrol car stationed at a mileage marker. Four minutes later, the car passes another patrolman at a marker that is five miles from the first. The driver notes that the speedometer registers 55 miles per hour. Can the highway patrol use the Mean Value Theorem to prove that the driver was exceeding the speed limit of 70 miles per hour at some time between the two mileage markers?

• • -=,A s(t) B ,1.

t:.o t ::. /tS

Let -t = elapsed time}i" hwrs a.fter ca..r

pQSses the ftrst pa.tro \ co..r:

S(-t) = d ,stance, ntil3 fr<'"' f, rs t Marker at time

Jt.

t =- � == _L tour �o 15

v��= S(-½)- S(o2+s--o

Yo.v;j = 15 m i/nrs

5 _.1_ IS

Yes, blJ MvT en.r wo.s trb.veli"9 a± '15 MP. h rl least DY\ee Or"

0 "f- M i /\, LJ"tttS

EX #4: Let h(x) = 5 -�, find all values of c in the openX

r : : . : : : -·-6 •. -)- . ···i······ ! · - - -+· ··Fh(xJ·--interval (1, 4) guaranteed by the Mean Value

Theorem.

h (4) - h(11 ;:: �4-1

I _!±_ x;,.

x.1;:;; 4-X-::: ± c:1

)(;!.

• •• C == J 8at1sf1es MV/ DY\l<X<4.

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I I I I I l I t I I I •

- ---•l••-•-•► • N -••♦•••••�••••••►••••••• • • ••

I I I • ' ' ' ' ' '

- .) .... :-----:····(41.tJ) ..... I J t I I I , I I t

---2 -- -: -- : -----:-----�------�-----�-----' ' ' + ' ' ' .

2 , 4 : 6 -- -:------�- ---!--·--�------�--- -�------

• I I I f

l I • I

I I I I

- -2 -- -·;------�-----:-----�------�----- -----' f I • . ' . . j t • I

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Topic 5.5: Using the Candidates Test to Determine Absolute (Global) Extrema

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EX #5: Use a calculator to find how many times the graph of y = x2 - 3 + sin(2x - 1) changes ii concavity on the interval [O, 2n].

So lve ( � - Lt sin ( Q x - )) =- 0 J

X I D < x < �lT)

X-;- l�n1r +lf-t � .

x= l�n1T + 57T -+fa';).

) I ;l,

n= o, 1, J,3r ..

X== o .. ,<c,18, ;,...:: J.&ogq, x== .3.9033) x= 4 .. 95os, x/ 1.0LJ�

4 -times

EX #6: Using a calculator, find the value( s) of x at which the gr.3.ph of y = ex x2 changes concavity. ii T:t-N.sP J f<E ex- CA5 Press MENU; �:5 (twice)

l/ x� -o.5851 X� -3. y I�� -�

;--�

To12tc �.7: !Jiing thi Se�ong 12erivgtive Test to Determine Extrema

Using the Second Derivative to find Extrema of a Function

Similarly, the SECOND DERIVATIVE is the FIRST DERIVATIVE of f'(x), so the same relationships that exist between f (x) and f' (x) must also exist between f' (x) and f" (x). We can use the SECOND

DERIVATIVE TEST as an alternate method for locating relative extrema.

If x = c is a critical number of f(x), then f' (c) = 0 orf' (c) does not exist. This makesx = c a "possible" maximum or minimum value of the function f(x).

f" (x) f(x) f'(x)

positive ,nu-easi'"j

negative

Concave. up

conco..ve rJown decre.o.s1°r) 9

neither possible po,ot of possfbte relative. lf\ffect,on he(e max/m1() here_

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EX #7: Use the second derivative test to find the relative extrema for each function. Justify.

A. g(x) = x4

- 8x3 - 5

9 /(x) = Lf-x 3_ 6llf x ;;,

4x 3 -�'-f x� = olf.X� ( X -�) =. 0X =- 0 J X = (o uiti·coJ

numbers

3//(x)= l�x..l -48x 3n( o) = I J(q.l _t.J8(0)

9q(o)� o 3 /I ( G, )-; f ;;l ( Cot - LJ.8 ( (o)

9'"Ceo) � 1J.f'+

B. f(x) = ½x - sinx on -2rr::; x::; n

+ �1(;<)= Si0x

-f (-J)" �1!- Si/\{--f.} f (--m: -{,- - Sil' (-f} f(lf/3)= -t- SiA(f)

x=c f(c) I" (c)

0 -5 -lefo (p - '-/37 posi·hve

Conclusion�.-

\) Local mi{) occ..vrs a± X="

Clr'd_ is - 431 b/c., �/"(b):>o .9 1 s concave u P·

�) No concLUS10() Clbove X= OSit1ce 3//(0) = o.

Conc.Jusions: t) LocQI mins ctr�

� -3 .. J.l-84 @ X-= - STr/3 a.ndz -o. 31.fJ @x= "o/3 s,nc.e.f // (x) > 0 a.nd f' i S C.Ol"Ca

up. �) Loca.l max value 1 s o.ooui

0 .. 34� @) X-;: -TrJ3 SiAce

-P /fx) <O o."d f is Clf\cave, 22

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EX #9: In the table at right, selected values for g(x), g' (x) and g" (x) are shown. The function y = g(x) is a twice-differentiable function and continuous on -4 5'.; x 5'.; 5.

X

-4

3

g(x) g' (x) g" (x)

0 -6 -1

4 0 -2

A. Explain why there must be a value x = c on ( -4, 5)such thatg(c) = 7. 5 8 3 0

q(-4)::-.0 _9(5)= �; S,oc.e 0<7< 8 li0d -f is corrr,nuouS on [- 4; 5], by 1vr iher� must be o. va.fue x=c. on (-4-,S) where 9Ce)=7.

B. Explain why there must be a value x = don (-4, 5) such that g' (d) = �

�/(d) = 9.Cs)-3(4) =__]_;since f is co�tinuous on E'+,5J5- (- lt) 9

and different,.o.bJe on (-4-,S)j by M VT there ,s o. valu�

ouch that 3/(d) = +. C. Does g(x) have a local maximum, local minimum or neither at x = 3? Justify.

3/(3)=0 c.r,-+icaJ nutr1ber O.r\d_ 3//(3)=-�. '30_3Cx) 1s COY'cove dowr1 and 9(3) is o.. local max1_01 um of 4 at x= .3 by JYl'd Derivo.tive Test- for 1::.xfre.mo..

D. Can concavity of g(x) be determined on the interval -4 < x < S? Why or why not?

No1 there i's not enot9 h i()furma:t, on to de1t?(m,11e CDf\cov·rty �V\ .- _4<X<5. We o(\l'-1 have ef\dpoints ard X = 3 �o 1()Hfl1fel

y ma0y po;nt5 are miss 111--9 to

est a._� I 1s h CDr1 cavi:fy

f"(x) > 0

f"(x) < 0

f"(x) = 0

SUMMARY OF FIRST AND SECOND DERIVATIVE TESTS

RELATING FUNCTION BEHAVIOR:

f'(x) > 0 f'(x) < 0 f'(x) = O

Increasing and Relative Minimum Concave Up and Concave Up

Increasing and Relative Maximum Concave Down and Concave Down

Increasing and Function is "smooth, level" and a possible Inflection Point

Decreasing and Concave Up

Decreasing and Concave Down

Decreasing and Inflection Point

' inflection point

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EX #6: The graph of y = f' (x) is shown below, on the interval [-8, 8]. If f (-8) = 5. On the same set of axes , sketch a possible graph of the function y = f (x).

f{\QX

( - g l 5) . . p6L

�/ - 0 +

X

-1-

-r -g "-s J'I -, \, 3 ?'- 8

+u I ..,. ..,

-f -g V

0 0 +-+-+ I - - - I

+ .f-

-.3 f'\ 1 u EX #7: The function f (x) is defined and differentiable on

the closed interval [-6, 6]. The graph of y = f' (x),

the derivative of/, consists of three line segments and a semicircle, shown in the figure at right. For each question below find the values for y = f(x) on the interval -6 < x < 6 and justify your reason.

f +-+ I 5V8

y =f'(x):

4 6

;{: -4 ; -:! ;

. .

A. Find the x-coordinate of each critical point for y = f (x).

-+fx)=O ; X=-4 Q0d X =- 0

. . . . ..... _, __ ·······�·····- ... }_., ; ' : -. ' . . . . . . - ... . . ' : : : . . . . . . ' . .

. ' . . . ' ... · ..... · ·····-···· ;··········�---··

B. Find the x-coordinate of each relative extrema of y = f (x) and label as a maximum or minimum.

At ?<� - Lf., ..f'/ changes_ s19n -f'rom ne3a+i'!e to posd-,ve. J so x:: -4 ,s o. r-ela.+,ve m1·A1mum.

C. Find the open interval(s) over which the function y = f (x) is increasing or decreasing.

Where -P ✓ex) >O J -9 is 10creas1n9 on (- Lf,o) u (o, �) W'riere +"()()<o

) f ,·cs decreas,n3 e,r, (-c,>-Lf)

D. Find the x-coordinate of each point of inflection for y = f (x).

Af x:4 & x=-c::), f/tx) cha.f\�S 019(\ from pos,+,ve to ne_ga-hve. At X= 0 J-f/r(x) cha'1(Jes s, gn f'rom negative to positive

f .,, ( -;)) = o o.n d f'/1' Cit) a n e & . f"(0) dneE. Find the interval(s) over which the function y = f(x) is concave up or concave down.

+/"(x) >O, f ,s concave up (- 6)-c2) u (o,4) -t11(x)<o1 -F t·s concave down (-J,o)u(L.J.,<o)

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EX #8: Let h(x) be a function that is continuous on the interval (0, 6]. The function his twice differentiable except at x = 3, where the derivative of h does not exist. The function

and its derivatives have the properties indicated in the table below. Use the table to sketch a graph ofy = h(x).

X 0 O<x<2

h(x) 3 Positive

h'(x) -3 Negative

h" (x) -2 Negative

2

0

-3

0

y --4 --·

2<x<3

Negative

Negative

Positive

. ' ' '

3

-3

DNE

ONE

I I t 0

-- - ,, --·- �- ____ ., - - . .. _j_ - - - .... - ----' 1 I I I o t • I •

' • t I I

--3 --- �-- .. -�-----�----�----�-----' I I I I

: : I ; : --2 ---: ---:9---T----�-----r--- ---

• I I l

--1 ___ _: - _1 ____ _! _____ _., ____ �---- ... -- -

; : ; ' ; X

·-1--2

'

--3 -- � ----�- - ----�----+----� - --' I I 0

I l • I I

--4 •--+----�----�-----�- --+----!-----. ' ' . ' ' ' . ' . . ' ' ' . ' ' '

EX #9: The function f is defined and differentiable on the closed interval [-6, 5). The graph of y = f' (x) the derivative off, is shown in the figure below and has horizontal tangent lines at x = -3 and x = 2.

A. Find the x-coordinate(s) of each relative extrema on-6 < x < 5. Identify as a max or min. Justify.

3<x<4

Negative

Positive

Negative

(-6, 9)

(-5.0) mat: x= -5> )(=4 j .f/ chat"\ges

513n -from c+) +o c-)min: X.:: o, +

., changes from E-) to ( +)B. Find the x-coordinate of each point of inflection for

y = f (x). Justify.

f . .,., cha('\9es s, 9n .Prom (-) �o (-+)a+ X= -3 Qnd -From (-+") +o C-·)

a.t X.=�-

C. Letg(x) = x3 - f(x). Findg'(-3).

3'(x) = 3 )( i - f /(x)

3 ✓(-3)� 3(-3)� - f/(-3)

9 ✓(-3) � J1-(-�)

(-3, -6)

5 5<x<6

0 Positive

1 Positive

0 Positive

.V

y=f'{x)

(2. 7)

(4, 0)

X

(5. -3)

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