Guys here's the math module! Due on Thursday. 12 NN. (Aug 18, 2011).

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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·
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PROBlEM1. Verify IVT given the function fdefinedby f(x)=4+3x-x2 in the il~rval 2 s 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 such
that 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,
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f(c)=4+3c-C2=1 q c2-3c-3=O q c
3±J21
2
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
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t
-4.5 s X s 3 for k = 3.
PROBlEM2. Verify IVT given f(x)=~ 25_X2
SOlUTlON2. To verify the intermediate vakJe theorem ifk =
il the interval
tt1l we need to find a number c in the intelVal [-4.5, 3)
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such thatf(c) = 3.
The function f is continuous on its domain M 01ll.
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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=±4
However, 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 ) = @.
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PROBlEM3. Verify IVT given the function fdefined by f(x)=-- in the in1erva1 -3 s X s 1for k = ~. f:
x+2 >,
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~:. SOlUTION3. To verify the intermediate vakJe theorem if k = ~ we need b find a number c in the interval [(20), @J
such 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).
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Thus, NT cannot guarantee the existenoe of a c between -3 and 1 so that f(c) = ~.
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ToiDustratefurther, f(-3)={H} =t:- f(1)=@.However, f(c)=--=- q c=6 "
c+2 2
,.' But then C = 6 (2: [-3,1].REMEMBER, WE CANNOT USE IVT WHEN THE FUNCTION IS DISCONTINUOUS!!!
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PROBLEM4. Use IZT to show that y = x3 - 4x 2 + X + 3 has a root between 1 and 2.
SOlUTION4. At x = 1: y = 1 - 4 + 1 + 3 = U§l > 0
Atx=2:y=8-16+2+3=@ <0
Since polynomials are continuous l?!l then IZT guanm~s that there exists a number C E ml such that
rnn
P(c) = 0; That is, there is a between 1 and 2. REMEMBER,IZT GUARANTEES THE EXISTENCE OF A ROOT
BUT DOES NOT PROVIDE A VEHICLE TO IDENTIFY THE PARTICUlAR VAlUE OF THE ROOT. ("; )
CLASS ID: NAME: DUE: Aug. 1812NN
(pigeonhole clo secretary)
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END OF REQUIREMENT
dlmgutierrez