Let f ( x) be continuous on a closed interval [a , b] and differentiable on the open interval (a , b), If f (a) = f (b) , then there is at least one point c âˆˆ (a, b) where fâ€™(c) =0.

**Rolleâ€™s Theorem**

Let *f* ( *x*) be continuous on a closed interval [*a* , *b*]
and differentiable on the open interval (*a*
, *b*)

If *f* (*a*) = *f* (*b*) , then there is at least one point *c* âˆˆ (*a*, *b*) where *f*â€™(*c*)
=0.

Geometrically
this means that if the tangent is moving along the curve starting at* x *= a towards as in Figâ€‰7.2 *x* =*
b *then there exists a c âˆˆ (
*a, b*) at which the tangent is parallel
to the* x *-axis.

** **

**Example 7.19**

Compute
the value of '*c* ' satisfied by the
Rolleâ€™s theorem for the function

*f*(*x*) =
*x* ^{2} (1 âˆ’
*x*)^{2} , *x* âˆˆ[0,1] .

**Solution**

Observe
that, *f*(0) =
0 =
*f* (1), *f* (*x*) is continuous in
the interval [0,1] and is differentiable in (0,1) . Now,

*f *â€²(*
x*)* *=* *2*x *(1âˆ’* x*)(1âˆ’* *2*x*)*
*.

Therefore,* f *â€²(c) = 0 gives c = 0,1 , and 1/2

which â‡’ c = 1/2 âˆˆ(0,1) .

** **

**Example 7.20**

Find the
value in the interval (1/2, 2) satisfied by the Rolle's theorem for the
function *f*(x) =* x *+ 1/*x* ,* x *âˆˆ [1/2 , 2] .

**Solution**

We have,
*f (x)* is continuous in [1/2 , 2] and
differentiable in (1/2 , 2) with* f*(1/2)* *= 5/2 =* f *(2) . By the Rolleâ€™s theorem there must exist a value c âˆˆ (1/2 , 2) such that,

* f*
(c) = 1 â€“ 1/c^{2} = 0 â‡’
c^{2} = 1 gives c = Â±1. As 1âˆˆ (1/2 , 2) , we choose c = 1.

**Example 7.21**

Compute
the value of '*c* ' satisfied by
Rolleâ€™s theorem for the function in the interval
[2, 3].

**Solution**

Observe
that, *f*(2) =
0 =
*f* (3) and *f* ( *x*) is continuous in
the interval [2, 3] and differentiable in (2, 3) . Now,

Observe
that âˆ’
âˆš6 âˆ‰(2, 3) and hence *c* = +
âˆš6 satisfies the Rolleâ€™s theorem.

Rolleâ€™s
theorem can also be used to compute the number of roots of an algebraic
equation in an interval without actually solving the equation.

** **

Without
actually solving show that the equation *x*^{4}
+
2*x*^{3} âˆ’
2 =
0 has only one real root in the interval (0,1) .

**Solution**

*Let f(x)= x*^{4}* + 2x ^{3} âˆ’ 2. Then f *(

*f *â€²(*
x*)* *= 4*x*^{3} +
6*x*^{2} . If *f* â€²(*x*)
=
0 then,

2*x*^{2} (2*x* + 3) = 0 .

Therefore,* x *= 0,âˆ’ 3/2 but 0, âˆ’ 3/2 âˆ‰(0,1) .

Thus, *fâ€™(x)* > 0, âˆ€x âˆˆ(0,1) .

Hence by
the Rolleâ€™s theorem there do not exist *a*
, *b* âˆˆ(0,1) such that, *f* (*a*)
=
0 =
*f* (*b*) . Therefore the equation *f*
(*x*) = 0 cannot have two roots in the
interval (0,1) . But, *f* (0) = âˆ’2
<
0 and *f* (1) =
1 >
0 tells us the curve *y* =
*f* ( *x*) crosses the *x* -axis
between 0 and 1 only once by the Intermediate value theorem. Therefore the
equation *x*^{4} +
2*x*^{3} âˆ’
2 =
0 has only one real root in the interval (0,1) .

As an
application of the Rolleâ€™s theorem we have the following,

** **

Prove
using the Rolleâ€™s theorem that between any two distinct real zeros of the
polynomial

a_{0}x^{n}
+ a_{n-1}x^{nâˆ’1} + â€¦.. + a_{1}*x *+ a_{0}

there is
a zero of the polynomial

na_{n}*x*^{n âˆ’1} + ( n âˆ’1)a_{nâˆ’1}x^{nâˆ’2}
+ â€¦ + a_{1}.

**Solution**

Let P (
x) = a_{0}x^{n} + a_{n-1}x^{nâˆ’1} + â€¦.. + a_{1}*x *+ a_{0}. Let Î± < Î² be two
real zeros of P ( x) . Therefore, P (Î± ) = P(Î² ) = 0 . Since P ( x) is
continuous in [Î± , Î² ] and differentiable in (Î± , Î² ) by an application of
Rolleâ€™s theorem there exists Î³ âˆˆ (Î±,
Î² ) such that P (Î³ ) = 0. Since,

*P*â€²(*x*) = na_{n}*x*^{n âˆ’1} + ( n âˆ’1)a_{nâˆ’1}x^{nâˆ’2}
+ â€¦ + a_{1}.

which
completes the proof.

** **

Prove
that there is a zero of the polynomial, 2*x*^{3}
âˆ’
9*x*^{2} âˆ’11*x* +12 in the interval (2, 7) given that
and 7 are the zeros of the polynomial *x*^{4}
âˆ’
6*x*^{3} âˆ’11*x*^{2} +
24*x* + 28 .

**Solution**

Applying
the above example 7.23 with

*P*(*x*)* *=* x *^{4}* *âˆ’* *6*x*^{3}* *âˆ’* *11*x*^{2}* *+* *24*x *+* *28,* **Î±** *=* *2,* **Î²** *=* *7

and
observing

*P*â€™( x) / 2 = 2x^{3} âˆ’ 9x^{2} âˆ’ 11x +12 = Q(x)
, (say).

This
implies that there is a zero of the polynomial *Q* ( *x*) in the interval
(2, 7) .

For
verification,

*Q*(2) = 16âˆ’36âˆ’22+12 = 28âˆ’58
=âˆ’30<0

*Q*(7) = 686 âˆ’ 441âˆ’ 77 +12 = 698 âˆ’ 518 =180 > 0

From
this we may see that there is a zero of the polynomial *Q* ( *x*) in the interval
(2, 7) .

** **

There
are functions for which Rolleâ€™s theorem may not be applicable.

(1) For
the function *f* ( *x*) = | *x* |, *x* âˆˆ[âˆ’1,1] Rolleâ€™s theorem is not
applicable, even though *f *(âˆ’1)* *=* *1*
*=* f *(1)* *because* f *(* x*)*
*is not differentiable at* x *=* *0* *.

(2) For
the function,

even
though *f* (0) =
*f* (1) = 1 , Rolle's theorem is not
applicable because the function *f* ( *x*) is not continuous at *x* = 0 .

(3) For
the function *f (x)* = sin* x *,*
x *âˆˆ [0, Ï€/2] Rolleâ€™s theorem is not
applicable, even though *f (x)* is
continuous in the closed interval [0, Ï€/2] and differentiable in the open interval
(0, Ï€/2) because, 0 =* f *(0) â‰ * f*(Ï€/2)* *= 1.

If *f* ( *x*)
is continuous in the closed interval [*a*
, *b*] and differentiable in the open
interval (*a* , *b*) and even if *f* ( *a*) â‰ *f*
(*b*) then the Rolleâ€™s theorem can be
generalised as follows.

** **

If *f* is continuous on
a closed interval [*a* , *b*] , and *c* is any number between *f*
(*a*) and *f* (*b*) inclusive, then
there is at least one number *x* in the
closed interval [*a* , *b*] , such that *f* ( *x*) = *c* .

Tags : Mean Value Theorem | Mathematics , 12th Maths : UNIT 7 : Applications of Differential Calculus

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12th Maths : UNIT 7 : Applications of Differential Calculus : Rolleâ€™s Theorem | Mean Value Theorem | Mathematics

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