Theorems(IVT,EVT,andMVT) - South Hadley Public Schools · 2019. 12. 18. · 15. (calculatorallowed)...

Theorems (IVT, EVT, and MVT)

Students should be able to apply and have a geometric understanding of the following:e Intermediate Value Theorem

e Mean Value Theorem for derivatives

e Extreme Value Theorem

Multiple Choice

1. (calculator not allowed) ( UU 4 theorems apply)

If f is continuous for a < x < b and differentiable for a < x < b, which ofthe following could be

false?

(A) f(c)=AOLing some c suchthata<c<b.~J Mvva

~)’) ‘(c)=0 for some c such thata<c<b.

has a minimum value on a< x <b. T &EN T

(D) /has amaximum valueona<x<b. T EVTb

(E) I J(x) dx exists.

~ rrr2. (calculator not allowed)

The function f is defined on the closed interval [2,4] and /(2)=/(3)=/(4). On the

open interval (2,4), J is continuousandstrictly decreasing. Which of the following

statementsis true? Monotoni® —no turning poms

(ayy attains neither a minimum value nor a maximum valueonthe closed interval [2,4]. —

(B) (f attains a minimum value but doesnot attain a maximum value on the closed

interval [2,4].

(C) /f attains a maximum valuebut does not attain a minimum valueon the closed

interval [2,4].

(D) / attains both a minimum value and a maximum value onthe closed interval [2,4].

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3. (calculator not allowed)

2

Let f bea function with first derivative defined by f'(x)=

that f(1)=7 and f(5)=11. What value of x in the openinterval (1,5) satisfies the

conclusion of the Mean Value Theorem for f on the closedinterval [1, 5 | ?

for x > 0. It is known

:(A) 1 (B) f° (Cc) 10 C0) iD

oS. Wade 4

4. (calculator not allowed) Yrs

X 0

8 I(x) 1 k 2

wwaie feeimeanern

> in the table above. The equation f(x) = ; musthaveat least two solutionsin the

interval[0, 2] if k= L Z Vig-

we) + (1 0) 2 @

sit5. (calculator not allowed)

The function f is continuous onthe closed interval[0, 2] and hasvalues that are given

Let g bea continuousfunction on the closed interval [0, 1]. Let g(0)= l and g(l)=0.

Whichof the following is NOT necessarily true?

Tru(A) There exists a numberh in [0, 1] such that g(/) 2 g(x) for all x in [0, 1). Kr

e

(B) Forallaand bin [0, 1], ifa=b,then g(a)=g(0). ©

(C) There exists a numberh in [0, 1] such that g(h) = Lo2

(D) ‘There exists anumberh in [0, 1] such that g(/) = -. WO

(E) For all hin the open interval (0, 1), lim g(x) = g(h).

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1.aas


(xIf f(x) =sin(= then there exists a numberc in the interval = <<- that satisfies

the conclusion of the Mean Value Theorem. Whichofthe following could be c?

20 = v24) 3 L(y) + con() £(82)= .

7 : ~_@) 4($)\> F(c)

4 uc(%)= 07 2 grr $a

Oy C0$20 %. 9,21* 2

X= Ty avy om


Whichofthe following theorems maybeapplied to the graph below,

y =|x-3|+, b> 0, overthe interval [2, 4]?

AY 4

Vx

I. Mean Value Theorem

IL. Intermediate Value Theorem

Ill. Extreme Value Theorem

(A) Ionly (B) only’ (C) Ilonly (E) I, I, and II (D) Il and III only


(a2 Nb lor! = t8. (calculator not allowed) ‘lpy)=

The function f is defined by f(x) =4x* —5x+1. The application of the Mean Value

Theorem to fon the interval 0 < x < 2 guarantees the existence of a value c , where

0<c<2 suchthat f(d)=

(A) 1 (C) 7 (D) 8

f(2\-*lo) a= ec 22-0 2


A function of f is continuous on the closed interval [ 2,5] with f(2)=17 and f(5)=17.

Whichofthe following additional conditions guarantees that there is a number c in the open

interval (2,5) such that f(c)=02.

(A) No additional conditions are necessary

hasa relative extremum on the open interval (2,5).

ifferentiable on the open interval (2,5).

(D) i f(x)dx exists


F(a) 3 4 9 13

fia 0 1 1 2

The table abovegives valuesof a differentiable function f andits derivatives at

selected values of x. If » is the function given by A(x) =f(2x), whichofthe following

statements mustbe true? h' ¥) - 2 fa \ (2x)

() A isincreasingon 2<x<4. KOT rewssandry | _ _

iv (i)

_

There exists c , where 0 < ec < 4, such that h(c)=12.~ h 0) — 3) \oe=

« (ll) There exists c , where 0) < c < 2, such that h(c)=3.A ((2-h4. |to)

7 > 12 be- II only 2-D

Land OF

©) II and " only


11. (calculator allowed)

The function fis continuous for —2 < x <1 anddifferentiable for —-2 < x <1. If

f(2) =andf()) = 4, which of the following statements could be false?

(A)_ There exists c, where —2<c<], such that So) =0.T vt £1) _f(-2) _

here exists c, where —-2<c<], such that f’(c) =0.

(C) There exists c, where —-2<c< 1, such that f(c)=3. 1 \

(D) There exists c, where -2<c<], such that f'(C)=3. 7 nN

(E) There exists c, where —2 < c <1 suchthat SO2zf@ TT

for all x on the closed interval -2 < x <1.

12. (calculator not allowed) y Vy \x 0 2 5 9 11

2(x) 1 2.8 1.7 1 3.4

The table above showsselcted values of a continuous function g. For OS x<11, whatis the

fewest possible numberoftimes g(x) =2? IW

(A) One (B) Two . (D) Four


Free Response

13.

14.

(calculator not allowed)

t 0 1 2 3 4 5 6(minutes)

Ci) 0 5.3 8.8 11.2 12.8 13.8 14.5(ounces)

Hot wateris dripping through a coffeemaker,filling ata upswith coffee. The amount ofcoffee in the cup at time 4, 0 <t <6, is given by a differentiable action C, where ¢ ismeasured in minutes. Selected values of C(t), measuréd in ounces,are given in the tableabove. MVT oypp|(10s

Is there atime t, 2 <t¢<4,at which C'(t)=2? Justify your answer.

“lA = Cun ele for Some tin (2,4)Ln) =

Ifa)\o 12.9- 8.8 _ _ ~ (ll -2 bu MVTC' (+) ate fa 5 C'(+)=2 by

(calculator not allowed)

t

(minutes) 0 2 . a ia

v4) 0 100 40 -120 -150(meters/minute)

Train A runs back and forth on an east-west se 2measured in meters per minute,is give a differentiable

measured in minutes. Selected values for v,(z) are given in the table above.

oad track. Train A’s velocity,ion v(t), where timet is

(b) Do the datain the table support the conclusionthat train A’s velocity is —100 meters perminute at some time ¢ with 5<+¢ <8? Give a reason for your answer.

V(5)= 40

MA(Y\=-120 _

— 00ike‘bebe“yte)ond v08), Trwrofee,

ley, INT NV(t)\= -(00 for Some tin (5,8),



Graph of f

A continuous function f is defined on the closed interval —4 < x < 6. The graph of if

consists of a line segment and a curvethat is tangentto the x-axis at x = 3 , as shown in the

figure above. Onthe interval 0 < x < 6, the function J is twice differentiable, with

f'"(x) >0.

(c) Is there a value of a, —4 < a < 6, for which the Mean Value Theorem,appliedto the

interval la, 6|, guarantees a value c, a < c <6, at which f'(c) = 3 Justify your

answer. C0) _ £ (tb) —F C0!) _ =

(pb —&

=f). 1 Let aes, FQ)O oonO

See MT oypplies

| ~“D- +j\(-f(sh. +r * 3 on (3,6).

16. (calculator not allowed) y

(3.4)

Graphof2”

Let g be a continuous function with 8(2) =5. The graph of the piecewise-linear function g, the

derivative ofg, is shown above for —3 < x <7.

(d) Find the average rate of change of g’(x) on the interval —3 < x <7. Does the Mean

Value Theorem applied on the interval —3 < x <7 guarantee a valueofc, for

—-3<c< suchthat_g"(c)torate of change? Why or whynot?a oo

§ || (=a) IeLkJ-4 ~

| V7 + + udh Q Value for CcMyT does not quaranter Sue |

beCOuUR "O'R not 0 fLerenbhable on (3,7)sConch©2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org

ncTETTET

esinen


x f@) £O) 8) 2’@6 4 2 5

2 9 2 3 13 10 -4 4 24 "| 3 6 7

The functions fand g aré differentiable for a

above gives valuesofthe functions and

is given by A(x) = fi(e(x))- 6.

real numbers,andg is strictly increasing. The table

their first derivatives at selected values ofx. The function 4

(a) Explain why there must be a value r for 1 < r < 3 such that A(r) =—5.

Nay= F(acn-b = £(2)-b= 4-4 > ht) =3

h(3)= £(gtay-b=+) -0 =--b»hl(aje-Fa ravd bla). By WT ANU! ts Some IAS hes betoeon lt) dna (3), By | 5 (3) suo aed

(b) Explain why there must be a value c for 1 < c < 3 such that h'(c) =

hic) =-S%,

Apely Mv T

h'(e) = hla hl) fy ge C10 (1,3)

3

Wle}= 322 = -B 5 hw -5| 2

—_



Let f 1atwice-differentiablo-fn ction such that f(2) =5 and i(5) =2. Let g be the function

given by g(x) = f(f(x).

(a) Explain whythere mustbe a value c for 2 < c < 5 such that f’(c)=-l.

Apay MVT

Oro) = flsh=tl2e) for some C19 (2,9S-L

fe) <= 2-5 =~-A4

(b) Show that 2’(2)=2"(5). Use this result to explain why there must be a value k for

2<k<5 such that g"(k)=0.

d= F(a)FO)e\- Ei (e(ay\Fi2) = FS) F(z)a(S) = $1 (#5))- Fs) = FC FG) oe

\'(2\= 4'(5) By Rolle's Theorem "EB=6 for So

(d) Let a=fyeewhywr anebea valent 2 J < 5 such thal " (2S)

h(r) = 0.

h(2)=flz)-2 = S-2 +23 hé2)=3

hts)=f{s)-S = 2-S = -3 = hlS)=-3

0) (es between bh(2) deol h(S). By INT

rere 1§ Same © IA (2,5) Sutthat

hUr)= O.

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Theorems(IVT,EVT,andMVT) - South Hadley Public Schools · 2019. 12. 18. · 15. (calculatorallowed)...

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