Mean Value Theorems

8/12/2019 Mean Value Theorems

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(iii) f l ( x ) - 3 ( ~ ~ + 3 ~ + 1 ) ~ ( 2 x + 3 ) .f ( x ) = 6 ~ ~ + 3 ~ + 1 ) 2 r + 3 ) ~ + 6 ~ ~ + 3 ~ + 1 ) ~

El3a5v = - dvA -2 = 6 4 x 2 ~ = 64.

- e t e1. (i) sinhx = 2 so-(sinhx) =r 2 cosh xsinhx @sh2x-sinh2x(ii) tanh x = , -sech2xI 12. cosh-'x = ymeancnshy = x,sosinhy9 land2r r sinhmsh2y - ~ i n h ~ ~ + l s o s i n b ~ ~x 2 - l s o s i ~ l h ~d

d 13 ( i) r n ( s i n f i ) = c o s ( f i ) f11 2 r(ii) og, ( x 2 + 1 p - -2 Xlog, e =2dxTxdxT x 2 +

El6Ay ~ ( X + A ~ ) ~ - ~ ( X + A X ) - ~ X ~ - ~ X& A ~ + 2 ( A x ) ~ - 5 A x

( 4 ~ - 5 ) A x + 2 ( A X ) ~dy f l ( x ) A x ( 4 r - 5 ) h nSoAy = dy+2(Ax)2Forx = 5, Ax = 0.1,Ay = 20 -5 ) 0.1 ) 2 0.1 )2 = 1.5 0.02 = 1.52.

El7~y - X + A X ) ~ - ~ . ~ + A X ) ~ + ~ ~ + A ~ ) - I - X ~ + ~ ~ ~ - ~ X += ~ x ~ - ~ ~ + ~ ) A x + ~ x - ~ + A K ) A K ) ~j y X ) A x + ~ ( A x ) ~where = 3x-3+Ax.Meredy = f l (x )A x = (3 x2 -6r+ 3)Ax.If x = 2,Ax = 0.02thenAy = (12-12+3)(0 .02) = 0.06.

Limit Continuity andDitTerentiability

- NowHence


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f ( a ) o r f ( b ) s o t h a t f ( c ) = M f ( a ) * c af ( c ) = M f ( b ) b.

This means that c lies in the oper interval a,b. We shall now show that c is the pointwhere ( c ) = 0 ince f ( c ) is an absolute maximum off (x), we have

then( x ) f s 0 whenever Ax >0Ax

and( ) -() 0 whenever Ax c 0Ax

It is given thatf (x) exists for all x E a, b ).Therefore,

c + A x ) - f ( ~ )rim f = f c) s 0A x - O X

The only possibility for0 s (c) s 0 s f (c) = 0.Consequently we have shown the existence of at least one point c E ( a, b ) wheref (x) vanishes. There may exist more than one such point.The conclusion of Rolle s theorem also holds true if we replace the conditionf (a) = f (b) by the conditionf (a) = f (b) = 0 eeping other conditions the same.Rolle s theorem has a very simple geometrical interpretation. If the graph of a

Mean Value l eorems

function is an unbroken curve intersecting thex axis at the points a and b and ifthe curve has a tangent at every point except, possibly, at the end points, then theremust be at least one point (c , (c) ) on the curve different from the end points atwhich the tangent is parallel to the x-axis (Figure 3.1 (a) and 3.1 (b)).

Example :The polynomial functionf (x) = x3 -xis continuous and differentiable for all real x. If we take a = -1 and = 1 ehave f (-1) = 0 = f (1). Therefore the conditions of Rolle s theorem are satisfiedon [-1 , 1 1 Thus there must be at least a number c such that -1 c c 1 andf (c) = 3c2- 1 = 0 n fact we have two values


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Both themots rebetween -1 and 1.

Thehct ion

iszematx Oandx land* ubtsontheopenintewal 0,1).Butthefunction isnot continuous on [0 , I ] .Tbe end pointx 1 is a point ofdiscontinuity. Therefore Rollers theorem is not applicableon he interval [0,11You can observe hat* different um i c m dy/ 1 ) on 0, l) .

ElCan Rolle s theorembe appliedto each of the following functions? Find c in caseit can be applied.(a) f x ) - six? x on he internal [ 0, x ]@ f x ) - 2 + 4 on [-2,2]c) f x ) - s i n x + c o s x on(d) f x ) - 2-2xon[0,1].

Letf x ) - a bx c Ifp and q are two real numbers such thatf@ f q),prove that

3 3 LAGRANGE S MEAN-VALUE THEOREMLetus oonsidera fun tion x)which satisfies all that conditions of Rolle s theorem exceptthe condition, f a ) = b).Then the conclusion of the theoremneed not hold trueFigure3.2).That is there need notbe a p6int c,f (c ) ) o he graph where the tangent is


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parallel to the x-axis. However, there appears to be a point where the tangent is parallel tothe chord that joins the points a, f a ) ) and b, f b ) of the graph = (x). Ageneralization of Rolle s Theorem, called Mean-Value Theorem, says that this will alwayshappen, if the curve y f x) is continuous on [ a, b ] and differentiable on a,b).Theorem 2 : The Mean Value Theorem

Letf a, b ]- be a function such that(1) f is continuous on [ a, b ](2) is differentiable on aJbThen there exists at least one point c a, b ) such that

We shall prove the mean-value theorem by constructing a function that satisfiesthe conditions of Rolle s theorem.We know that

(b) -f (a)b-a

is the slope of the chord joining P : a, a ))and Q : b,f b )) and its equation isx-a b-a

x -a ) where xJy)y(x)-f(a)+ b-ais a point on the line PQ.

Define a function a, b ]- by G (x) f (x) x ) (the vertical distance betweenthe curvey =f (x) and the chord PQ at an arbitrary point x of [aJb . That is,

4 f ( 4 f ( 4 - L H (x.-a)b-aWe see that

(1) is continuous over [ a, b ]2) G = f (x)- exists at each point of a, b ) andb-a

(3) G (a) G (b) 0.Hence Rolle s theorem applies to G (x), and there exists a number c a, b ) for whichG (c) Othat is

f (b) f a)f (4 = b-o


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W e give below tw o equivalent forms of the above conclusion of the mean-value theorem:i) f b ) - f a ) - b - a ) f c ) , w h e r e a < c < b

( i i ) f l ( a + h ) - f ( a ) = hf ( a + B h) w he r e O < B < l .Example3 :

Le ty = nd -2 s 2.Find a number c such that the tangent to the curve at the point P :. c, f c ) ) ispan llel to the chord joiningA : - , -8 ) and B : 2 , 8 )We see that = .? satisfies the conditions of lagr ange s mean-value theorem.

ryThe slope of the tangent to y = s 3 2 .The slope ofAB is

By Lagrange s theorem we have a point c such that -2 c < 2 andf (c) = 3c2 = 4which implies that c here are two values of cd 5between 2 and 2 where the tangent to the curve y x3 is parallel to the chordAB.

Example 4 :Check whether Lagrange s formula is applicable to the function y = 1- Y3overthe interval [ 1 l ]

Solution :ll)e function is continuous in [ -1 , 1 ]Tlie derivativeY x) - exists at all non-zero points of I 1.1 and does not exist atx 0. Moreover. the sl ow ofA B is

2and y ' c) = -- c) for any finite c.Lagrange s formula does not hold in this case because the derivativef (x) fails toe x i s t a t a p o i n t o f ( - 1 , 1 ) , y = f x)doesnothaveatangentatx=O.

The following four important theorems arebased on Lagrange s mean-valuetheorem.


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Theorem :Let f x )be a continuous function over [ a,b ] and differentiable over q b . Iff x ) 0 for all x a, b ) then x ) s a constant function over [ q b

I f f x ) s non-constant and continuous over [ q b then there exist two distinctpoints x , and x2 of [ a,b ] such thatf x , ) f x J . Then by Lagrange s mean-valuetheorem on [ x,, x2 ] we get a point c such thatxl < c 0 there. If x , and + are twonumbers such that a < x , < x2 < b. Then x, ) < x2).

We consider the interval [ xl, x2 1 By Lagrange s theorem there exists a number csuch that xl < c < x2

f (4 f x2) f ~ 1 )x2 - 1 1It is given that x2 >xl andf c )> 0 Therefore, we get f x 2 ) x,) > or

Theorem6 :Let f x )be differentiable over a, b ), andf x ) x2).You can prove the Theorem on similar lines of Theorem5.Verify the Mean Value Theorem MVT) for the following functions:

(b) f x ) = h 3 on [ 0 , 2 ]


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ELetf be a function on [4 6 ] satisfying the assumptions ofMVT Let c be a pointguaranteed by theMVT Rove that if g x) = f x ) 1and h x) = f (x) 2 for allx in [ u,b 1, then th same point c satisfies gl(c)and

b-ahO h- hl(c)also.b-a

3 4 INCRE SING ND DECRE SING FUNCTIONSfunctionf x) is said to be increasing (or decreasing) over an intervalI iff (x,) c f (xz) (or

f (x,) >f (x2) ) whenever x,, I and x, 0 over a, b ) thenf (x) is increasing over a, b ).It can alsobe shown that iff (x)> 0 over a, b ) and iff (x) is continuous over [ a, b ) (or

a, b ] thenf (x) is increasing over [ a, b ) or a, b ] ). You can prove these results onyour own, by showing thatf (a) OIf xE (O ,2 ) then f (x)O

It follows thatf (x) is increasing over o O ] decreasing wer [ 0,2 ] andincreasing wer [ 2, ) can you draw the graph?)

Show that the functionf (x) = tan x increases on -n/2, n/2 ) and n/2,3n/2 ).


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EXExamine whether the functionf (x) = s increasing over [ 0, n/2 ).CQSX

L e t y = ~-fisinx,~sxsx.~indthevaluesofx~[~,x]wherey c~,~0a n d y l > 0.

3 5 CAUCHY S MEAN-VALUE THEOREMThis heorem is also called generalised m ean-value theorem for derivatives and is useful inevaluating indeterminate forms which you will be studying in the next unitTheorem 6

ff x) and g ( x )are umtinuous functions aver [ a, b ] and differentiable overa , ) and for x a, ), g x ) z 0 then there exists a number c such thata < c < b .and

Prod:First of all we shall show that

g ( a ) z g ( b ) if g ( x ) 0 for x E ( a , b ) .If we suppose g ( a ) g ( b ) hen by R olle s theorem we shall have at least onepoint c a , b ) such that g ( c ) = 0. Bu t it cannot happen since g (x) r 0 forevery x a , b ).Thus g ( a ) H g (b).We next w nsider the function

You can check that h x)satisfies all the conditions of Rolle s theorem. Therefo re,there exists a point c E (a , b ) such that h ( c ) = 0

You can note the following:1) If g (x) x , then Cauchy s theorem reduces toLagrange s hemem.2) We cann ot prove the Cauchy s theorem by applying Lagrange s theorem to

the num erator and deno rnimto r of the fraction separatelyg t b ) - g ( a )because by L agrange s theorem we shall get


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(b) - (a) ~1)- a(b)-g(a) - g(b)- g(a) g (c2)b -a

where a u c, < b and a < c2< b andc, need not be. equal to c This result is not thatof Cauchy.

Example6 :DO he functionsf (x) = and g (x) = 2ix atisfy the conditions of Cauchy stheorem in the interval [ 3 ] ?

SolutionOne of the conditions of the Cauchy s theorem is g (x) * or x a, b ). Here

3B ( 4 = = -3andb = 3.Wenotethatg(-3) = g(3),g(x)is2continuous over [ - 3,3 ] and gl(x) exists over - 3 , 3 ).Therefore by Rolle stheorem there is a number such that -3 < c < 3 and g l c ) = and the conditiong (x) + orx - 3 , 3 ) does not hold. Cauchy s theorem is not applicable here.

Example :Use Cauchy s theorem to show that x


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by su ch an approximation, so that you can im prove the result by adding more and m oreten ns of th e power series until the approximation is satisfactory. The general method forexpanding functions into power series is by m eans of Taylor s formula. We formulate theproblem a s follows. We qssume that the function y = f ( x )has derivatives up to n 1 )thorder, inclusive, in som e neighbourhood of the point x = a. We shall find a polynomialP, (x )of degree such that

d P . ( x ) = f ( a ) ..., , ( a ) = ( a ) ,(3 .1)

It is reasonable to accept that such a polynomial, in some sense, is close to the functionf (x ) .

P , ( x ) = c , + c , ( x - a ) + ~ ~ ( x - a ) ~ +.. c , ( x - a ) " .We shall determine the constants co, c l ,c ~ . . ., so that the conditions 3.1)will be satisfied.Then we obtain

f ( a ) = P,,a ) = C o

n . n - 1 ) .. .2 .1 c ,

co = f ( a ) , , = @ c2 r ( a c n - f ( n ' ( a )l 2 nx - a ) ( ~ - a ) ~ ( ~ - a ) ~ f " ( a )P, = f ( a )+ (a) + ( a ) ... n ( 3 .2 )

Let R, ( x )be the difference of values of the functionf x ) and the correspondingpolynomial constructed P, (x),Figure 3 4

Y

igurc 3 4

pproximation for (x ).Here R, ( x ) s called remainder. W e shall try now to find a suitable form of R, ( x ) n termsof some higher order derivative of f ( x )so that we can estim ate the error involved in thepolymial approxim ation of a function. The result is given in the following theorem which iscalled Taylor s Formula with a remainder.

Mean-Value heorems


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Theorem7 :(Taylor's Fornula withRemainder)If f (x), @),f" x), ...,f"i@) re mtinuow on he ose nterval [ a, b ] and i ff ( + ) (x) xists in the open n t e ~ da,b )then thereexists a point c a < c


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ThereforeK (b) = f (" ' (c),

for somec between a and b.The theorem is proved.

Corollary :It follows fiom the Taylor's theorem that :If f (x) has derivatives of all orders in a neighbourhood of the point x = a, thenfor each positive integer n and for each x

- a n (x-a)'"n ( a) + ( n + l ) f'" +" (c )

for some c between a and x.Wecanwritec = a+O(x-a )whereO


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y s i nx y(0) = 0y' m s x sin(:+x), yl(0) = sinz 12y" = -s inx = sin 2 . - + x yn(0) = s i n n = 0

y = sin n . x y (0) = s n ..aThen the Taylor s formula for sin x about x s

w he re 0 a c a x .W e note that if the function sin x is approximated by the polynom ial

x - - + - - ... ? sin n3 J nthe error involved is

for some c between and x.Then

To obtain the abov e inequality we have used the,,fact( s i n 0 1 1.

In the lim it, as n - =we get

Thesefore we can make the error as small as we like by taking sufficiently large n.Suab kind of approximate polynomials are generally used to find approximate values ofsine for 8 fixed.Tb e graphs of the function sin x and first three approximations:P (x) = x

are shown in Figure 35.W e ob serve that the approximation improves, if n is increased.Example8

Use M aclaurin s formula and expand the functiony = log (1 x ), defined in[0 1] n powers of x. Estimate the absolute error due to deleting the remainderR 4.


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Mean-Value bcoren~s

Solutionf 0) log, 1 = 0

f ( x ) n- ( n - I ) f ( n ) ( ~ ) ( - ~ ) ~ - l ( n - l )( x + l ln

Therefore the Maclaurin formula for log, ( 1 x ) is

whereO


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up (I) P(6) (1) .....y (1)where c is between 1and xWe obtain

p ( l ) 0, d ( 1 ) = 0 , ~ 1) - 0p 3) (1) 18,p 4) (1) 72,p (1) = 120p 6 ) (x) = 0,p 7) (x) 0,

Therefore,

= 3 ( ~ - 1 ) ~ + 3 ( x - l ) ~ + ( x - l ) ~ .This equality is exact and holds true for all x

E

UseMaclaurin's formula to establish the inequality os x 2/2

Write Taylor's formula of the functiony d about the point a = 1withremainderR (x).

erify the quadratic approximations 1 x 2 near x = a ~ ~ deterlni~~e1 xits accuracy for ( x .1.


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3.7 SUMM RY

In this u nit we havk1 proved Rolle s theorem and learned the geometrical meaning of it;2. used R olle s theorem t o prove Lagrange s mean-value theorem and Cauchy smean-value theorem and learned about their applications;3. derived a few conseq uence s of Lagrange s theorem;. derived the T aylor s formula with remainder for functions satisfying certainconditions and used it to write polynomial approximations for given func tions andestimated the error involved in such approximations.

3.8 SOLUTIONS / NSWERS

(a) f x ) = sin2x is continuous on [ 0, x 1 differentiable n0, n and f 0 ) = 0 = f x ) so Rolle \ theorem can be applied. Nowf ( c ) = 2 s i n c c o s c = Oforc = -2

@ f x ) = x + is continuous on [ - 2 2 1 differentiable in- 2 ) and f - ) = 22 = f 2 ) s o Rolle s theorem can be applied.N o w f f ( c ) - 2c = Oforc = 0.

(c) f x ) = sin x + cos x is continuous on 0 -- ifferentiable inI( 0 ,I 0 ) = 1 = z o Rolle slheorim cannot be used One may see

ith at f ( c ) = 3 c 2 - 2 = Oforc = - o u t o f w ~ i c ~ ~ i s i n ~ ~ , ~ ~ s othat f c ) = 0 may be true even if Rolle s theorem cannot be used.

E Take the interval [p, q 1. Then Rolle s theorem can be used forf ( X ) = x + x + c sincef ( x ) is continuous on [p , q 1 differentiable in ( p , q )a n d f ( p ) = f ( q ) . T h e n f f ( c ) = 2 a c b = O f o r s o m e c i n [ p , q ] . H e r ec = - ow consider the equationf x ) = x +bx +c = 0.Obviousl y if the2a


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-btwomotsarepandq,thenp+q = -sothatc = - ( p + q ) anda

(a) f x ) = tanx is continuousin1-0 4f l ( c ) = sec c = = .e. sec c = . Such a point will existX--0 i4i Xbecause osc = -for some between [ 0,- For this note that4

intermediate value theorem, some in1l -3 4 - which is inside the interval.) f l ( c ) - 4 c + 3 = 4'

1(c) .J c ) = 3c = - = 1 which means c = * both of which are in theinterval.E4

Wehavegl (x) = f l x ) f o r a l l x in a , b ) a n d g b ) - g a ) = f ( b ) - f ( a ) .Agamhl (x) f l x ) f o r a l l x i n a , b ) a n d h ; b ) - h a ) f ( b ) - f ( a ) . /

Sf x ) tan x meansf ; = sec2x > 0 for all x; so it is increasing everywhere,and in particularon the intervals given.

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increasing.E7

y x - f i s h , * l ( x ) = 1 - ~ c ~ . S o f ' ( x ) < O f o r

decreases for 0 to- nd its value decreases from 0 to -1 we have

3(a) Cauchy's MVT conditions are satisfied,f x ) -x -x, g x ) = x H 0

cancelled since c H 0.

b -aNow 1 a fi'V-P b 2 - a 2 s 0 C a s ]b - a


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Meaa Value lheorems


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UN T 4 APPLICATIONSStructure

4.1 IntroductionObjectives4.2 Applications to Geometry4 2 1 Geometrical Meaning of the Derivative

4 2 2 Equations of the Tangent and Normal a t a Point4 2 3 Angle of Intersections between Two urves4 2 4 Differential Coefficient of the Length of an Arc of y f (

4.3 Maxima and Minima4 3 1 Necessary Condition for a Maximum and Minimum4 3 2 Geometrical Interpretation4 3 3 Points wheref ( ) does not exist4 3 4 First Suflicient Condition for a Maximum or a Minimum-Change of Sign ofFirst Derivative4 3 5 Rule for Finding Maxima and Minima4 3 6 SecondSufticient Condition for Maximum and Minimum UseSeoondDerivative)4 3 7 Greatest and Least Values of a Fundion in Closed Interval4 3 8 Maxima and Minima Problems

4.4 Indeterminate Forms4 4 1 The Indeterminate Form O4 4 2 The ndeterminate Form /4 4 3 Further Indeterminate Forms

4.9 Summary

4 1 INTRODUCTIONIn Unit 2, we defined the derivative of a function-Westudied several important theoremslike Rolle' theorem, Mean value theorems etc. Differential Calculus is a very useful tool instudying pnctical problems. n this unit we shall study some applications of the derivativeof a function. In Section 4.2 we propose to study application of derivativesto geometry. In4.3 we shall discuss how derivatives can be used to determine the points where adifferentiable function has maxima and minima and how to solve practical problemsinvolving minimisation or maximisation of some functions.

bjectivesAfter reading this unit, you shouldbe able to

find the tangent and normal to the graph of a given function at given points,find the angleof intersection between two curves,find the curvature, radius, centre and circle of curvature of a curve,locite the points where a function has a maximum or a minimum,solve some problems when it is required to minimise or maximise a function, andto evaluate limits of functionswhich have indeterminateforms0/0, m/= 0.m- O, o. mO or IS.

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Mean Value Theorems