DERIVE for Windows version 5.06 DfW file saved on 30 Nov 2002
P(x):=x^3 - 3·x^2 + x + 3
Q(x):=x^2 - 4·x + 5
Q1(x):=x - 5
Q2(x):=(Q1(x) - Q1(-1))/(x + 1)
hCross:=APPROX(32258064516129031/1000000000000000000)
vCross:=APPROX(7407407407407407/200000000000000000)
ÿÿCTextObjè·ÿ0{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\qc\cf1\b\f0\fs24 The Remainder Theorem, Polynomial Expansion,
\par & Tangent Lines
\par \pard\b0
\par NOTE: \cf2 Before starting this activity, we recommend that you open a \f1 2D-\f0 plot window, return to this \f1 window\f0 , and choose the \f1 Window > \f0 Tile Vertically\f1 option\f0 . In this way you will be able to view the graphs you create while follow\f1 i\f0 ng the text.\cf1
\par
\par \b Background\b0
\par
\par \cf3\f1 Plot\f0 the graph of the following polynomial in the \f1 2D-\f0 plot window.\cf1\f2
\par }
ÿÿCExpnObjHÃÐÛ
Simp(User)
x^3-3*x^2+x+3€çè
ÿ÷{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1\fcharset0 Times New Roman;}{\f1\fmodern\fprq1 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Use the Set > Plot Range command in the 2D-plot window\f1 \f0 to adjust the window so that \f1 -2 \ul <\ulnone \i x \ul < \ulnone\i0 2 and -6 \ul <\ulnone \i y \ul\i0 < \ulnone 6. \f0 Plot the line in expression #2.\cf2\f2
\par }
€h°%
Simp(User)ú~j¼t“h?y=x+3€1èÉÿn{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Note that the graph of the above straight line is tangent to the graph of the polynomial at \i x\i0 = 0. Why? To answer this question first observe that the right hand side of the straight line equation is made up of the constant term and the linear term of the polynomial in expression #1. This in itself doesn't tell us much.
\par
\par Another fact is very important for realizing the significance of the above observation. In particular, we will see:
\par \i when x is close to 0, the constant term plus the linear term dominate the value of the polynomial.\cf2\i0 \cf1 First, we subtract off these two terms from the polynomial.\f1 \cf2 Plot \f0 the graph of the result\f1 (expression #3 below)\f0 . Restrict the \f1 plot range\f0 to -.1 \ul <\ulnone \i x \ul <\ulnone\i0 .1 and -.05 \ul <\ulnone \i y \ul\i0 < \ulnone .05.\f2
\par }
€8Õ€íUserð¿ x^3-3*x^2€ùèkÿ–{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Note that the graph of the polynomial in expression #3 lies entirely within this window. This means that the sum of the constant term and linear term will never differ from the value of the polynomial by more than .05 for any \i x\i0 -value within the interval -.1 \ul <\ulnone \i x \ul <\ulnone\i0 .1.
\par
\par Another way to see why the lower degree terms dominate the polynomial is to look at the table generated by the following command.\f1
\par }
€8w¨Userð¿*APPROX(VECTOR([x,x^2,x^3],x,-0.9,0.9,0.1))€8›èWSimp(#4){®Gáz„?ÿM[[-0.9,0.81,-0.729],[-0.8,0.64,-0.512],[-0.7,0.49,-0.343],[-0.6,0.36,-0.216],[-0.5,0.25,-0.125],[-0.4,0.16,-0.064],[-0.3,0.09,-0.027],[-0.2,0.04,-0.008],[-0.1,0.01,-0.001],[0,0,0],[0.1,0.01,0.001],[0.2,0.04,0.008],[0.3,0.09,0.027],[0.4,0.16,0.064],[0.5,0.25,0.125],[0.6,0.36,0.216],[0.7,0.49,0.343],[0.8,0.64,0.512],[0.9,0.81,0.729]]€cèÌÿr{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 The table illustrates the fact that
\par \i If -1 < x < 1\i0 , \i then x is greater than any of its higher powers. In fact, the higher the power the greater the dominance of x.
\par
\par \i0 All of this supports the fact, but does not prove it, that the tangent line to the graph of a polynomial at \i x\i0 = 0 can be found by the equation: \i y = b*x + a \i0 where \i a\i0 is the constant term and \i b*x\i0 is the linear term.
\par
\par \ul Exercise\ulnone
\par \cf2 1) Choose a fourth degree polynomial. Draw its graph and choose an appropriate \f1 plot range\f0 for viewing the polynomial. Draw the graph of the tangent line to the polynomial at \i x\i0 = 0. Find \f1 an \f0 interval for \i x\i0 such that the values for \i y\i0 on the tangent line are within .05 of the value of the polynomial.
\par \cf1
\par On\f1 e\f0 side effect of our observation is that if we wish to make a quick approximation to the value of a polynomial at a "small" value for \i x\i0 , we can do it by using the last two terms of the polynomial - the constant and linear term.
\par
\par \b Points Other Than Zero\b0
\par
\par Suppose we want to approximate a polynomial or find the tangent line at a point other than 0. How can we do it? For example, suppose we are interested in values near \i x\i0 = -1. \cf2 \f1 Simplify\f0 the expression:\cf1\f2
\par }
€8ØhðUserð¿(x+1)^3-6*(x+1)^2+10*(x+1)-2€üè"ÿÁ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 It is just the polynomial for expression #1 written in a different way\f1 ?\f0 \cf2 Let's plot the straight line made up from the last two parts of the polynomial\cf1 .\f2
\par }
€8.È:Userð¿y=10*(x+1)-2€FèÂÿÉ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
{\colortbl ;\red0\green0\blue0;\red255\green0\blue0;}
\viewkind4\uc1\pard\cf1\f0\fs24 In this instance we have the graph of the tangent line to the polynomial at \i x\i0 = -1. The explanation is basically the same as before: when \i x\i0 is close to -1, then (\i x\i0 + 1) dominates all higher powers of (\i x\i0 + 1).
\par
\par \ul Exercise\ulnone
\par \cf2 2) Explain the above statement.\cf1
\par
\par The interesting question is how did we know that expression #5 was equivalent to our original polynomial? Furthermore, can this be done for any value of \i x\i0 ? The answer to both of these questions can be found using a very important theorem in Algebra. It is called\f1 :\f0
\par
\par \b The Remainder Theorem.\b0
\par
\par This theorem states:\i
\par If P(x) is a polynomial of degree n \ul >\ulnone 1 and a is any real number, then there is a polynomial, Q(x), of degree n-1 and a number, r, such that
\par P(x) = Q(x)(x - a) + r.
\par Furthermore, r = P(a).
\par
\par \i0 How do we find the polynomial \i Q\i0 (\i x\i0 )? The theorem tells us. \cf2 Solve the equation for \i Q\i0 (\i x\i0 ) by \f1 simplifying\f0 the following Derive steps.\cf1 We use the polynomial from expression #1 and the point \i a\i0 = -1.\f2
\par }
€8ÎæUserð¿P(x):=x^3-3*x^2+x+3€8òèUserð¿ Q(x):=(P(x)-P(-1))/(x-(-1))€"ènÿÿ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Thus, we see for any Polynomial, \i P\i0 (\i x\i0 ), division of (\i P\i0 (\i x\i0 )-\i P\i0 (\i a\i0 )) by (\i x -\i0 \i a\i0 ) will always result in a polynomial. (Note that if \i P\i0 (\i x\i0 ) is a constant polynomial the result will be another constant polynomial, 0.) To see the result given in the statement of the theorem, highlight \i Q\i0 (\i x\i0 ) in expression #9 and evaluate it. Then highlight \i P\i0 (-1) in that result and evaluate it. This will give the right hand side of the expression in the statement of the Remainder Theorem.\f1
\par }
€8zà†Userð¿
Q(x)*(x-(-1))+P(-1)€’è¥ÿx{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
{\colortbl ;\red0\green0\blue0;\red255\green0\blue0;}
\viewkind4\uc1\pard\cf1\f0\fs24 Since \i Q\i0 (\i x\i0 ) is a polynomial, we can continue this process. \cf2\f1 Simplify\cf1\f0 :\f2
\par }
€8±ðÕUserð¿Q1(x):=(Q(x)-Q(-1))/(x-(-1))€áèS ÿs{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\ul\f0\fs24 Exercise\ulnone
\par \cf2 3)\cf1 \cf2 Using the polynomial you graphed in Exercise 1) find\f1 \i\f0 Q\i0 (\i x\i0 ) and \i Q\i0 1(\i x\i0 ) for this polynomial using a value for \i a\i0 that is within the range of \i x\i0 -values in your graphing window.
\par
\par \cf1\f1 Now we h\f0 ighlight the \i Q\i0 (\i x\i0 ) in expression #9 and choose the \ul S\ulnone ubexpression substitution option from the \ul S\ulnone implify menu. Typ\f1 ing\f0 (\i Q\i0 1(\i x\i0 )(x-(-1))+\i Q\i0 (1)) in the substitution box and click\f1 ing\f0 the OK button \f1 we\f0 obtain\cf3 :\f2
\par }
€8_ €k Sub(#10')ð¿%(Q1(x)*(x-(-1))+Q(-1))*(x-(-1))+P(-1)€w èŠ ÿ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Multiply this by hand and we have the expression\f1
\par }
€8– €® Userð¿
&Q1(x)*(x-(-1))^2+Q1(-1)*(x-(-1))+P(-1)€º èà ÿÙ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Next \f1 we \f0 evaluate \i Q\i0 1(\i x\i0 ), \i Q\i0 (\i -1\i0 ), \i Q\i0 (\i x\i0 ), and \i P\i0 (-1) in\i \i0 expression #12. This will require four partial evaluations. The result is shown in expression #13.\f2
\par }
€8ì P
Userð¿(x-5)*(x-(-1))^2+10*(x-(-1))-2€
èô
ÿÈ{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
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\viewkind4\uc1\pard\cf1\f0\fs24 Compare this expression with expression #5!\cf2 The last two terms are the expression for the right hand side of the tangent line equation in expression #6. The remaining terms are multiplied by (x - (-1))^2. Thus, we have anticipated the following result.
\par \i If P(x) is a polynomial then the tangent line to the graph of P(x) at x = a is given by the expression
\par Q(a)(x - a) + P(a)
\par where Q(a) is the polynomial (P(x) - P(a))/(x - a).\i0
\par The proof of this general result is obtained by applying the Remaider Theorem twice as we did in our specific example.
\par
\par \ul Exercises\ulnone
\par \cf1 4) Using your polynomial from Exercise (1 and the results from Exercise (3. Find the equation of the tangent line to the graph of the polynomial at \i x = a\i0 for the value of \i a\i0 that you selected in exercise (3.
\par 5) Create a polynomial \i Q\i0 2(\i x\i0 ) defined by\f1 :\f2
\par }
€8$Userð¿Q2(x):=(Q1(x)-Q1(-1))/(x-(-1))€0èIÿa{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fmodern\fprq1 Times New Roman;}{\f1\fmodern\fprq1\fcharset0 Times New Roman;}{\f2\fmodern\fprq1\fcharset2 DfW5 Printer;}}
{\colortbl ;\red255\green0\blue0;\red0\green0\blue0;}
\viewkind4\uc1\pard\cf1\f0\fs24 and finish the expansion of \i P\i0 (\i x\i0 ) in terms of powers of (x - (-1)). Your answer should be equiv\f1 a\f0 lent to expression #5.
\par \cf2
\par We have used a theoretical result to find a geometric object, namely the tangent line to the graph of a function. Places of interest on the graph are those where the tangent line is parallel to the \i x\i0 -axis. That is, those points where the slope is equal to 0. At these points the graph of a polynomial will have either a peak or a valley (local maximum or local minimum). Using this fact we can solve problems for finding the "best" value provided the value is given as a polynomial expression. We provide an example problem for you to work on.
\par
\par \ul Exercise\ulnone
\par \cf1 6)\cf2 \cf1 Suppose you are given a piece of wire 20 inches long. You are told to cut it in two pieces. One is to be bent into the shape of a circle and the other into the shape of a square. It is well known that if you wish to enclose the maximum are\f1 a\f0 with your wire, then you don't cut it at all and bend it into the shape of a circle. However, suppose you wish to enclose the \ul minimum amount of area\ulnone with the wire and the shape restrictions. Is the solution to bend it all in the shape of a square?
\par \cf2
\par \fs20 Carl Leinbach: leinbach@gettysburg.edu\f2\fs24
\par }
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