Mathematical Methods for Physics and Engineering : A Comprehensive Guide

2.1 DIFFERENTIATION

and hence the difference between the curve and the line is

h(x)=f(x)−g(x)=f(x)−f(a)−(x−a)

f(c)−f(a)

c−a

.

Since the curve and the line intersect atAandC,h(x) = 0 at both of these points.

Hence, by an application of Rolle’s theorem,h′(x) = 0 for at least one pointb

betweenAandC. Differentiating our expression forh(x), we find

h′(x)=f′(x)−

f(c)−f(a)

c−a

,

and hence atb,whereh′(x)=0,

f′(b)=

f(c)−f(a)

c−a

.

Applications of Rolle’s theorem and the mean value theorem

Since the validity of Rolle’s theorem is intuitively obvious, given the conditions

imposed onf(x), it will not be surprising that the problems that can be solved

by applications of the theorem alone are relatively simple ones. Nevertheless we

will illustrate it with the following example.

What semi-quantitative results can be deduced by applying Rolle’s theorem to the fol-

lowing functionsf(x),withaandcchosen so thatf(a)=f(c)=0?(i)sinx,(ii) cosx,

(iii)x^2 − 3 x+2,(iv)x^2 +7x+3,(v) 2x^3 − 9 x^2 − 24 x+k.

(i) If the consecutive values ofxthat make sinx=0areα 1 ,α 2 ,...(actuallyx=nπ,for
any integern) then Rolle’s theorem implies that the derivative of sinx,namelycosx,has
at least one zero lying between each pair of valuesαiandαi+1.
(ii) In an exactly similar way, we conclude that the derivative of cosx,namely−sinx,
has at least one zero lying between consecutive pairs of zeros of cosx.Thesetwore-
sults taken together (but neither separately) imply that sinxand cosxhave interleaving
zeros.
(iii) Forf(x)=x^2 − 3 x+2,f(a)=f(c)=0ifaandcare taken as 1 and 2 respectively.
Rolle’s theorem then implies thatf′(x)=2x−3=0hasasolutionx=bwithbin the
range 1<b<2. This is obviously so, sinceb=3/2.
(iv) Withf(x)=x^2 +7x+ 3, the theorem tells us that if there are two roots of
x^2 +7x+ 3 = 0 then they have the root off′(x)=2x+ 7 = 0 lying between them. Thus if
there are any (real) roots ofx^2 +7x+ 3 = 0 then they lie one on either side ofx=− 7 /2.
Theactualrootsare(− 7 ±

√

37)/ 2.

(v) Iff(x)=2x^3 − 9 x^2 − 24 x+kthenf′(x) = 0 is the equation 6x^2 − 18 x−24 = 0,
which has solutionsx=−1andx=4.Consequently,ifα 1 andα 2 are two different roots
off(x) = 0 then at least one of−1 and 4 must lie in the open intervalα 1 toα 2 .If,asis
the case for a certain range of values ofk,f(x) = 0 has three roots,α 1 ,α 2 andα 3 ,then
α 1 <− 1 <α 2 < 4 <α 3.