T,;,~ ..r':~\.."~

••'!:I.:'i1}'J.{:;"

. '"

'i:.-,

,.'~:.

'.",

,.'

.',.'.

MATH 5 MODULE 4 (2HW4)~TERMED~TEVALUETHEOREM

INSTRUCTIONS:Do not write anything on this page. List your answers a~r PROBLEM 4. FiRin the missing inbrmation in

numbers with parentheses ill·

PROBlEM1. Verify IVT given the function fdefinedby f(x)=4+3x-x2 in the il~rval 2s X s 5 fork = 1.

SOlUTION1. To verify the in~rmediate vakJe theorem ifk = 1 we need to find a nurroer c i1 the intelVall.{1}, WI suchthat f(c) = ro.Because f is a polynomial function, it is continuous on its domail Mi. Henoe, it is continuous on the closed intelVal [2,5)

Sinre f(2) = m =t:- f(5) = lID , IVT guarantees that there is a nurrber c between m and @ such that f(c) = 1. That is,

3±J21f(c)=4+3c-C2=1 q c2-3c-3=O q c2

However, m is an extraneous oolution because this number is outside the interval [2, 5]. Therefore, we accept only the

number 1m wh<h is in lhe desired illeJValand f ( 3+f21 ) =1 It

II

PROBlEM2. Verify IVT given f(x)=~ 25_X2 il the interval -4.5s X s 3 for k = 3.

SOlUTlON2. To verify the intermediate vakJe theorem ifk = tt1l we need to find a number c in the intelVal [-4.5, 3)such thatf(c) = 3.

The function f is continuous on its domain M 01ll.

Sinoo f(-4.5) = tl.4l =t:- f(3) = @, IVT guarantees that flare is a nurroerc between -4.5 and 3 sum that f(c) = (OO.That is, f(C)=~ 25-c2 =3 q 9=25-c2 q c=±4However, tl1l is an extraneous oolution because this number is outside the interval [-4.5, 3]. Therefore, we acoept only

the number {!§lwhich is in the desired interval and f (-4 ) = @. rf:I,

f:>,

4PROBlEM3. Verify IVT given the function fdefined by f(x)=-- in the in1erva1 -3 s X s 1for k = ~.

x+2

SOlUTION3. To verify the intermediate vakJe theorem if k = ~ we need b find a number c in the interval [(20), @Jsuch that f(c) = tm.Because f is a rational function, it is continuous on its domain @. Henoe, it is discontinuous on the interval [-3, 1).

Thus, NT cannot guarantee the existenoe of a c between -3 and 1 so that f(c) = ~.

4 1ToiDustratefurther, f(-3)={H} =t:- f(1)=@.However, f(c)=--=- q c=6

c+2 2But then C = 6 (2: [-3,1].REMEMBER, WE CANNOT USE IVT WHEN THE FUNCTION IS DISCONTINUOUS!!!

,I

"

fdlmgutielTezII'

I·

PROBLEM4. Use IZT to show that y = x3 - 4x 2 + X + 3 has a root between 1 and 2.

:",

Atx=2:y=8-16+2+3=@ <0

SOlUTION4. At x = 1:y = 1 - 4 + 1 +3 = U§l > 0

CLASS ID: NAME: DUE: Aug. 1812NN(pigeonhole clo secretary)

1. 16.

2. 17.

3. 18.

4. 19.

5. 20.

6. 21.

7. 22.

8. 23.

9. 24.

10. 25.

11. 26.

12. 27.

13. 28.

14. 29.

15. 30.

Since polynomials are continuous l?!l then IZT guanm~s that there exists a number C E ml such thatP(c) = 0; That is, there is a rnn between 1 and 2. REMEMBER,IZT GUARANTEES THE EXISTENCE OF A ROOTBUT DOES NOT PROVIDE A VEHICLE TO IDENTIFY THE PARTICUlAR VAlUE OF THE ROOT. ("; )

I,.

I..I'"

. .I

END OF REQUIREMENTdlmgutierrez