Intermediate Value Theorem, Rolle’s Theorem and Mean …

1 Intermediate Value Theorem , rolle s Theoremand Mean Value TheoremFebruary 21, 2014In many problems, you are asked to show that something exists, but are notrequired to give a specific example or formula for the answer. Often in this sortof problem, trying to produce a formula or specific example will be following three theorems are all powerful because they guarantee theexistence of certain numbers without giving specific 1( Intermediate Value Thoerem).Iffis a continuous function onthe closed interval[a,b], and ifdis betweenf(a)andf(b), then there is a numberc [a,b]withf(c) = an example, letf(x) = cos(x) x.

2 Sincef(0) = 1 andf( ) = 1 ,there must be a numbertbetween 0 and withf(t) = 0 (sotsatisfies cos(t) =t).It is not hard to get a decimal approximation totbut there is no simple formulafortusing standard 2( rolle s Theorem ).Supposefis continuous on[a,b]and differen-tiable on(a,b), and suppose thatf(a) =f(b). Then there is a numberc [a,b]withf (c) = 3(The Mean Value Theorem ).Supposefis continuous on[a,b]anddifferentiable on(a,b). Then there is a numberc [a,b]withf(b) f(a)b a=f (c).Problems1. Suppose thatfis continuous on [0,1] andf(0) =f(1).

3 Letnbe anynatural number. Prove that there is some numberxso thatf(x) =f(x+1n). (x) =f(x) f(x+ 1/n). Consider the set of numbersS={f(0),f(1/n),f(2/n),..,f(1)}. Letkbe such thatf(k/n) is the1largest number inS. Suppose thatk6= 0 andk6=n. Theng(k/n) =f(k/n) f((k+ 1)/n) 0, andg((k 1)/n) =f((k 1)/n) f(k/n) 0. By the IVT, there isc [(k 1)/n,k/n] withg(c) = 0, so thatf(c) f(c+ 1/n) = 0, orf(c) =f(c+ 1/n) as , if the largest number inSisf(0) =f(1), then the same argumentworks withkchosen so thatf(k/n) is the minimum number inS.

4 Andnote that iff(0) is both the largest and smallest number inS, then theyare all the same andf(0) =f(1/n).2. Given any two triangles in the plane, show that there is one line thatbisects both of the trianglesT1andT2. LetCbe the point which is thecenter of mass ofT1. For [0,2 ], let` be the oriented line throughCwhose positive direction makes an angle with the horizontal (a picturehere would really help). Every` bisects the first letf( ) be the portion of area of the second triangle which liesto the left of the oriented line`.

5 The functionfis continuous, andsatisfiesf( ) +f( + ) = Area(T2).In particular,f(0) +f( ) = Area(T2).Iff(0) = Area(T2)/2, we are done, and`0bisects both triangles. Iff(0)<Area(T2)/2 thenf( )>Area(T2)/2 and by the Intermediate valuetheorem, there is a withf( ) = Area(T2)/2, so that` bisects bothtriangles. Finally, the casef(0)>Area(T2)/2 hasf( )<Area(T2)/2 andagain by the IVT we get a line bisecting both A hiker begins a backpacking trip at 6am on Saturday morning, arrivingat camp at 6pm that evening. The next day, the hiker returns on the sametrail leaving at 6am in the morning and finishing at 6pm.

6 Show that thereis some place on the trail that the hiker visited at the same time of dayboth coming and (t) be the hiker s distance from the trailhead on the uphilltrip as a function of time, andd(t) be the hiker s distance from the trail-head on the downhill trip as a function of time. Letf(t) =d(t) u(t).Nowf(6am)>0 andf(6pm)<0, so by the IVT there is a timecwithf(c) = 0, which meansd(c) =u(c), so the hiker ais at the same locationon the trail on both trips at Leta,b, andcbe real numbers. Show that the equation4ax3+ 3bx2+ 2cx=a+b+calways has a root between 0 and (x) =ax4+bx3+cx2.

7 Thenf(0) = 0 andf(1) =a+b+ the mean Value Theorem , there is anx0between 0 and 1 withf (x0) =a+b+c, so 4ax30+ 3bx20+ 2cx0=a+b+cas Prove thatx3 3x+chas at most one root in [0,1], no matter whatcmay (x) =x3 3x+c. Iff(x) has rootsaandbon [0,1] thenby rolle s Theorem , there isc (a,b) withf (c) = 0. Butf (x) = 3x2 3is not zero for anyc (0,1), a contradiction, sofhas at most one root in[0,1].6. Fornany positive integer andx,yreal numbers, find all solutions to(xn+yn) = (x+y) only solutions arey= 0 orx= 0 or, whennis odd,x= y.

8 Fixy, and letf(x) =xn+yn (x+y)n. Computef (x) =n(xn 1 (x+y)n 1). Iff (x) = 0 thenxn 1= (x+y)n 1and takingn 1-th roots gives the solutionx=x+y, which works for anyxas longasy= 0. Whennis odd, there is the additional solution x=x+yorx= even. We havef(0) = 0, and if there is any otherx0withf(x0) = 0, then by rolle s Theorem , there is somecbetween 0 andx0withf (c) = 0, which can only happen wheny= 0. We have shown the onlysolutions arey= 0 orx= 0 odd. We havef(0) = 0 andf( y) = 0. If there is a thirdsolutionx0withf(x0) = 0 then by rolle s Theorem , there are two distinctsolutions forf (x) = 0, which can only happen wheny= 0.

9 We haveshown the only solutions arex= 0,x= y, ory= 0 Prove that for 0 a < b < /2,b acos2(a)<tanb tana <b acos2(b).8. Suppose at timet= 0, a particle is at rest. At timet= 1, the particle isat rest 1 unit from its starting position. Prove that at some moment theparticle s accelleration was For which real numberskdoes there exist a continuous real valued functionfsatisfyingf(f(x)) =kx9for all realx? solution exists if and only ifk 0. Ifk 0, letf(x) =k1 (f(x)) =kx9as desired. Now supposek <0, and suppose thereexistsfwithf(x) =kx9for , claimf(0) = 0.

10 We havef(f(0)) =k09= 0, sof(0) =f(f(f(0))) =k(f(0)) (0) = 0 or 1 =k(f(0))8, but the latter is impossible for negativek, sof(0) = , we produce ac6= 0 withf(c) = 0. Considerf(1), which may bezero, positive, or negative. Iff(1) = 0, letc= 1. Iff(1)>0, then leta= 1 andb=f(1). Thenf(a) =f(1)>0 andf(b) =f(f(1)) =k <0, so by the IVT there is acbetweenaandbwithf(c) = 0 sinceaandbare both positive. Iff(1)<0, then leta=f(1)<0 andb=f(f(1)) =k <0. Thenf(a) =f(f(1)) =k <0, andf(b) =f(f(a)) =ka9>0, so by theIVT, there iscbetweenaandbwithf(c) = 0 sinceaandbare both , we have a contradiction, sincekc9=f(f(c)) =f(0) = 0 impliesc= 0.