Unit 5: Intermediate value theorem

1 INTRODUCTION TO CALCULUSMATH 1 AUnit 5: Intermediate value solutions tog(x) =h(x) is equivalent to find solutionsf(x) =g(x) h(x) = :Iff(a) = 0, thenais called arootoff. Forf(x) = sin(x)for example, there are roots atx= 0,x= . are a few examplesExample:Find the roots off(x) = 4x+ : we setf(x) = 0 and solve forx. In this case 4x+ 6 = 0 and sox= 3 :Find the roots off(x) =x2+ 2x+ : Becausef(x) = (x+ 1)2the function has a root atx= 1. In generalx2+bx+chas :Find the roots off(x) = (x 2)(x+ 6)(x+ 3).Answer:Since thepolynomial is factored already, it is easy to see the rootsx= 2,x= 6,x= :f(x) = 12 +x 13x2 x3+x4. Find the roots off. There is no try (12 = 3 4 is a hint). We seex= 1,x= 3,x= 4,x= 1 are the :The functionf(x) = exp(x) does not have any :The functionf(x) = log|x|= ln|x|has rootsx= 1 andx= :f(x) = 2x 16 has the rootx= value theorem of continuous on the interval [a,b]andf(a),f(b) have different signs, then there is a root offin (a,b).

2 Proof is constructive: we can assumef(a)<0 andf(b)>0. The othercase is similar. Look atc= (a+b)/2. Iff(c)<0, then take [c,b] as the newinterval, otherwise, take [a,c]. We get a new root problem on a smaller the procedure. Afternsteps, the search is narrowed to an interval [un,vn] ofMATH 1 Alength 2 n(b a). Continuity assures thatf(un) f(vn) 0 and thatf(un),f(vn)have different signs. Both point sequencesun,vnconverge to a root :Verify that the functionf(x) =x17 x3+x5+ 5x7+ sin(x) has a function goes to + forx and to forx . We havefor examplef(10000)>0 andf( 1000000)<0. Use the :There is a solution to the equationxx= 10. Solution: forx= 1 we havexx= 1<10 forx= 10 we havexx= 1010>10.

3 Apply the Intermediate value theoremto the functionf(x) =xx :Earth is a point on the earth, where bothtemperature and pressure agree with the temperature andpressure on the Draw an arbitrary meridian through the poles and letf(x) be the temperatureon that circle, wherexis the polar angle. Define the functiong(x) =f(x) f(x+ ). Ifgis zero on the north pole, we have found our point. If not,g(x) has different signs onthe north and south pole. By the Intermediate value theorem , there exists therefore anx, whereg(x) = 0 and sof(x) =f(x+ ). For every meridian there is a latitude valuel(y) for which the temperature works. Define nowh(y) =l(y) l(y+ ). This functionis continuous. Start with the meridian 0.

4 Ifh(0) = 0 we have found our point. If not,thenh(0) andh( ) have different signs. By the Intermediate value theorem again,hhas a root. There, both temperature and pressure agree with the antipode :Wobbly Table an arbitrary floor, asquare table can be turned so that it does not wobbleany The 4 legs ABCD are located on a square in a plane. Letxbe the angle of thelineACwith with a coordinate axes if we look from above. Givenx, the table can bepositioneduniquely: the center of ABCD is on thez-axes, the legsABCare on thefloor andACpoints in the directionx. Letf(x) denote the height of the fourth legDfrom the ground. If we find an anglexsuch thatf(x) = 0, we have a position whereall four legs are on the ground.

5 Assumef(0) is positive. (f(0)<0 is similar.) Tilt thetable around the line AC so that the two legs B,D have the same vertical distancehfrom the ground. Now translate the table down byh. This does not change the anglexnor the center of the table. The two previously hovering legsBDnow touch theground and the two othersACare below. Now rotate aroundBDso that the thirdlegCis on the ground. The rotations and lowering procedures have not changed theINTRODUCTION TO CALCULUS location of the center of the table nor the direction. This position is the same as if wehad turned the table by /2. Thereforef( /2)<0. The Intermediate value theoremassures thatfhas a root between 0 and :The last application is a preparation for the derivative whichwill be introduced next week.

6 Lets callDf(x) = (f(x+h) f(x))/htheh-derivativeoff. We will study it more in the next :Lets call a pointp, whereDf(x) = 0 ah-critical call a pointaalocal maximumiff(a) f(x) in an open intervalcontaininga. Alocal minimumis a pointa, wheref(a) f(x).Example:As in the first lecture, we can check that the functionf(x) =x(x h)(x 2h) has the derivativeDf(x) = 3x(x h). With the notation [x]3=x(x h)(x 2h)and [x]2=x(x h), we haveD[x]3= 3[x]2. Since [x]2has exactly two roots 0,h, thefunction [x]3has 2 critical :More generally, for [x]n+1=x(x h)(x 2h) (x nh) we can computeD[x]n+1= (n+ 1)D[x]n. Because [x]nhas exactlynroots, the function [x]n+1hasexactlyncritical points. KeepD[x]n=n[x]n 1in mind!

7 For the usual derivative it will be trued/dxxn=nxn :You can verified that exph(x) = (1 +h)x/hsatisfiesDexph(x) = exph(x).Because this function has no roots and the derivative is the function itself, the functionhas noh-critical (x) = exp(x) following is an application of the Intermediate value theorem and also providesa constructive proof of theBolzano extremal value theoremwhich we will see s maximum theoremIffis continuous and hasf(a) =f(b) =f(a+h),thenfhas either a local maximum or local minimum inside the open interval (a,b). argument is to split the interval [a,b] into two [a,c] and [c,b] of the samelength. Now,f(c) f(a) andf(b) f(c) have different sign so thatg(x) =f(x+h/2) f(x) has different signsg(a) andg(c) at the end points.

8 By the intermediatevalue theorem there is a root ofgin [a,c] and therefore a pointxin [a,c] wheref(x) =f(x+h/2). This gives a new interval [a1,b1] of half the size where the situationf(a1) =f(b1) holds. Continuing like this we get a nested sequence of intervals [an,bn]which have size 2 nh. The limiting point is a maximum or minimum h-critical points in the limith 0 will become critical point . Buth-criticalpoints make sense for all continuous functions. We will need more regularity to takethe limith 0. The function needs to bedifferentiable. We proved now that anyh-critical point leads to either a minimum or 1 AHomeworkProblem :Find the roots for 72 54x+ 35x2+ 15x3 3x4 are told that all roots are :Use the Intermediate value theorem to verify thatf(x) =x7 6x6+ 8 has at least two roots on [ 2,2].

9 Problem :The Queen s gambit features two fine actors AnyaTaylor-Joy and Thomas Brodie-Sangster (both sharing expressive wideeyes). Anya s height is 170 cm, Thomas height is 178 cm. Anya was bornApril 16, 1996, Thomas was born on May 16, 1990. Anya s and Thomasnet worth are both estimated to be 3 ) Can you argue that there was a moment when Anya s height is exactlyhalf of Thomas height?b) Can you argue that there was a moment when Anya s age was exactlyhalf the age of Thomas?c) Can you argue that there as a moment when Anya s fortune was exactlyhalf of Thomas fortune?Argue with the Intermediate value theorem or note a scenario where thestatement is :Argue why there is a solution toa) 5 sin(x) =x, b) exp(7x) =x, c) sin(x) = ) Why does the following argument not work:The functionf(x) = 1/cos(x) satisfiesf(0) = 1 andf( ) = 1.

10 Thereexists therefore a pointxwheref(x) = ) Does the functionf(x) =x+log|log|x||have a root somewhere? Arguewith the Intermediate value :a) Leth= 1/2. Find ah-critical point for the functionf(x) =|x|. As defined in the text we look for a point for which [f(x+h) f(x)]/h= ) Verify that for anyh >0, the functionf(x) =x3has noh-criticalpoint. There is nox, where [f(x+h) f(x)]/h= 0 is Knill, Math 1a, Harvard College, Spring 2021