Rolle’s Theorem
     Statement
     Let a function f:[a,b]R be such that

     (i) f is continuous on [a,b],
     (ii) f is differentiable on (a,b), and
     (iii) f(a)=f(b).


Then there exists at least one point c(a,b) such that f   )=0.




 If f crosses the axis twice, somewhere between the two
crossings, the function is flat. The accurate statement of this
``obvious'' observation is Rolle's Theorem.
Interpretation of Rolle’s Theorem
Geometrical Interpretation:
If the function y=f(x) has a graph which is a continuous curve on
[a,b], and the curve has a tangent at every point of (a,b), and
f(a)=f(b), then there must exist at least one point c in (a,b) such that
the tangent to the curve at (c,f(c)) is parallel to the x-axis.
Algebraic interpretation:
If the function f(x) satisfies all the conditions of Rolle’s Theorem,
then between two zeros a and b of f(x) there exists at least one zero
of f (x) [i.e., between two roots a and b of f(x) =0 there must exist at
least one root of f (x)=0].




Immediate conclusion:
Between two consecutive roots of f (x)=0, there lies at most one root
of f(x)=0.

Important note:
The set of conditions in the Rolle’s Theorem is a set of sufficient
conditions.

The conditions are by no means necessary.

Illustrations:
   i) The function f defined in [0,1] as follows:
           f(x)=1, 0
              =2,              .
           The function f(x) satisfies none of the conditions of Rolle’s
           Theorem, yet f'(x)=0 for many points in [0,1].
ii) The function f defined in [-1,3] as follows:
         f(x)= +           .
         The function f(x) satisfies none of the conditions of Rolle’s
         Theorem, yet f'(x)=0 for many points in [-1,3],
         In fact f'(x)=0 for all x(0,1).

        The above two examples show that the conditions in the
        Rolle’s Theorem are not absolute necessity for f'(x) to be
        zero at some point in the concerned interval.

iii)    The function f defined in [-1,1] as follows:
        f(x)= .
        Here,
         f(x) is continuous on [-1,1],
        f(-1)=f(1),
        but f'(0) does not exists i.e. , f(x) is not differentiable on
        (-1,1).
        Also,
        f'(x) vanishes nowhere in (-1,1)-{0}.

        Failure of Rolle’s Theorem can be explained by the fact
        that     is not derivable in -1<x<1, though other
        conditions are satisfied.

iv)     The function f defined in [-1,1] as follows:
        f(x)=x3+5x-4.
        f(x) is continuous on [-1,1],
        f(x) is derivable on (-1,1),
        but f(-1)≠f(1).
        Also,
        f'(x) =3x2+5 vanishes nowhere in (-1,1).
        Failure of Rolle’s Theorem can be explained by the fact
        that                , though other conditions are satisfied.
v) The function f defined in [-1,1] as follows:
            f(x)=x2,      x=-1
                = 5x , -1<x<1
                = x2,      x=1

           Here,
           f(x) is continuous on (-1,1) and discontinuous at the points
           x=-1 and x=1.
            i.e., f(x) is not continuous on [-1,1],
            f(x) is derivable on (-1,1),
           and f(-1)=f(1).
           f'(x) vanishes nowhere in (-1,1).

           Failure of Rolle’s Theorem can be explained by the fact
           that f(x) is not continuous in -1    1, though other
           conditions are satisfied.



Examples (iii), (iv)& (v) show that if we drop any of the conditions in
Rolle’s Theorem then Rolle’s Theorem fails to be true.




           Important Note:
           If f(x) satisfies all the conditions of Rolle’s Theorem in
           [a,b] then the conclusion that f (c)=0 where a<c<b is
           assured, but if any of the conditions are violated then
           Rolle’s Theorem will not be necessarily true; it may still
           be true but the truth is not ensured.
Lagrange’s Mean Value Theorem
            (First Mean Value Theorem of Differential Calculus)


     Statement
     Let a function f:[a,b]R be such that

     i)     f is continuous on [a,b],
     ii)    f is differentiable on (a,b).


Then there exists at least one point c(a,b) such that




Another Form of Lagrange’s MVT:
If in the statement of the theorem, b is replaced by a+h, (h>0), then
the number between a and b may be written as a+θh, where
0<θ<1.

Then the theorem takes the following form:

     Let f:[a,a+h]R be such that

     iii)   f is continuous on [a,a+h],
     iv)    f is differentiable on (a,a+h).


Then there exists a real number θ lying between 0 and 1 such that
or, f(a+h)=f(a)+hf′(a+θh).

** Mean Value Theorem relates the mean rate of change
to instantaneous rate of the function.




Somewhere inside a chord, the tangent to f will be parallel to
the chord. The accurate statement of this common-sense
observation is the Mean Value Theorem
For any function that is continuous on [a, b] and differentiable
on (a, b) there exists some c in the interval (a, b) such that
the secant joining the endpoints of the interval [a, b] is
parallel to thetangent at c.




Geometrical Interpretation:
If the function y=f(x) has a graph which is a continuous curve on
[a,b], and the curve has a tangent at every point of (a,b),then there
exists at least one point c in (a,b) such that the tangent to the curve
at (c,f(c)) is parallel to the line segment joining the points (a,f(a)) and
(b,f(b)).




Remarks:
   i) The fraction             measures the mean (or average) rate of
            increase of the function in the interval [a,b] of length b-a.
            Hence the theorem expresses the fact that, under the
            conditions stated, the mean rate of increase in any
            interval [a,b] is equal to the actual rate of increase at
some point c within the interval (a,b). For instance, the
           mean velocity of a moving car in any interval of time is
           equal to the actual velocity at some instant within the
           interval. This justifies the name Mean Value Theorem.
 ii) Mean Value Theorem proposes that any differentiable function
           defined over an interval has a mean value at which a
           tangent line is parallel to the line joining the end points of
           the function’s graph on that interval.
 iii) Rolle’s Theorem is a particular case of Lagrange’s Mean Value
           Theorem. If f(a)=f(b) holds in addition to the two
           conditions of Mean Value Theorem then f(b)-f(a)=0 and
           consequently f (c)=0.




Important deductions:
   i)     Let f:[a,b]R be such that

          *f is continuous on [a,b],

          *f is differentiable on (a,b)

   And * f'(x)=0 for all x(a,b).

   Then f(x) is a constant function on [a,b].

   ii)    Let f:[a,b]R be such that

          *f is continuous on [a,b],

          *f is differentiable on (a,b)

   and * f'(x) 0 for all x(a,b).

   Then f(x) is a monotone increasing function on [a,b].
iii)   Let f:[a,b] R be such that

            *f is continuous on [a,b],

            *f is differentiable on (a,b)

     and * f'(x) 0 for all x(a,b).

     Then f(x) is a monotone decreasing function on [a,b].




    Cauchy’s Mean Value Theorem:
    (Second Mean Value Theorem of Differential Calculus)

     Statement
      Let the functions f:[a,b]R and g:[a,b]R be such that

     i)     both f,g are continuous on [a,b],
     ii)    both f,g are differentiable on (a,b), and
     iii)   f'(x)≠0 for x(a,b).


Then there exists at least one point (a,b) such that

                   .
Second Statement:
     Let the functions f:[a,a+h]R and g:[a,a+h]R be such that

     iv)   both f,g are continuous on [a,a+h],
     v)    both f,g are differentiable on (a,a+h), and
     vi)   f'(x)≠0 for x(a,a+h).


Then there exists at least one real number θ lying between 0 and 1
such that

                         .




Geometric Interpretation of Cauchy’s
MVT:
First Interpretation: The functions f and g can considered as
determining a curve in the plane by means of parametric equations
x=f(t),y=g(t),where atb. Cauchy’s MVT concludes that  a point
(f(),g())of the curve for some t= in (a,b) such that the slope of the
line segment joining the end points (f(a),g(a)) and (f(b),g(b)) of the
curve is equal to the slope      of the tangent to the curve at t=.
Second Interpretation: We may write




Hence, the ratio of the mean rates of increase of two functions in an
interval [a,b] is equal to the ratio of the actual rates of increases of
the functions at some point  within the interval (a,b).




REMARKS:
  i) Lagrange’s MVT can be deduced from Cauchy’s MVT by taking
            f(x)=x, x(a,b).
  ii) Rolle’s Theorem can be obtained from Cauchy’s MVT by letting
            f(x)=x and g(b)=g(a).
            [We have used Rolle’s Theorem to prove Cauchy’s MVT.]
iii) Both f and g satisfy the conditions of Lagrange’s MVT.
           Consequently there exist points c and d in (a,b) such that

                               and

          c and d are different points in (a,b) in general, and
          therefore a single point  in (a,b) may not be found to
          satisfy the conclusion of Cauchy’s MVT.

Some problems:
   Q1. Show that the equation 4x5+x3+7x-1=0 has exactly one
   real root.

   Q2. A function f is differentiable on [0,2] and
   f(0)=0,f(1)=1,f(2)=1.   Prove that f'(c)=0 for some c in (0,2).

   Q3. Prove that the equation        (x-1)3+(x-2)3+(x-3)3+(x-
   4)3=0      has only one real root.

   Q4.     Prove that between any two real roots of excosx+1=0
   there is at least one real root of exsinx+1=0.

   Q5.     Verify Lagrange’s MVT for the function f(x) =x(x-1)(x-2)
   in the interval [0, ].

    Q6.    Show that

                                        if 0<u<v

          deduce that

           <         <     .
Q7.     If in the Cauchy’s MVT we take f(x)= and g(x)=    ,
then prove that  is the arithmetic mean between a and b.
(Refer to the statement of Cauchy’s MVT).

Q8.      If f(x)= (x-a)m(x-b)n where m and n are positive integers,
show that c in Rolle’s theorem divides the segment a x b in
the ratio m:n.

Q9. Apply MVT to prove that            lies between 10 and
10.05.

Q10. Show that

                  , where 0<                 .

Mvtword

  • 1.
    Rolle’s Theorem Statement Let a function f:[a,b]R be such that (i) f is continuous on [a,b], (ii) f is differentiable on (a,b), and (iii) f(a)=f(b). Then there exists at least one point c(a,b) such that f )=0. If f crosses the axis twice, somewhere between the two crossings, the function is flat. The accurate statement of this ``obvious'' observation is Rolle's Theorem.
  • 2.
    Interpretation of Rolle’sTheorem Geometrical Interpretation: If the function y=f(x) has a graph which is a continuous curve on [a,b], and the curve has a tangent at every point of (a,b), and f(a)=f(b), then there must exist at least one point c in (a,b) such that the tangent to the curve at (c,f(c)) is parallel to the x-axis.
  • 3.
    Algebraic interpretation: If thefunction f(x) satisfies all the conditions of Rolle’s Theorem, then between two zeros a and b of f(x) there exists at least one zero of f (x) [i.e., between two roots a and b of f(x) =0 there must exist at least one root of f (x)=0]. Immediate conclusion: Between two consecutive roots of f (x)=0, there lies at most one root of f(x)=0. Important note: The set of conditions in the Rolle’s Theorem is a set of sufficient conditions. The conditions are by no means necessary. Illustrations: i) The function f defined in [0,1] as follows: f(x)=1, 0 =2, . The function f(x) satisfies none of the conditions of Rolle’s Theorem, yet f'(x)=0 for many points in [0,1].
  • 4.
    ii) The functionf defined in [-1,3] as follows: f(x)= + . The function f(x) satisfies none of the conditions of Rolle’s Theorem, yet f'(x)=0 for many points in [-1,3], In fact f'(x)=0 for all x(0,1). The above two examples show that the conditions in the Rolle’s Theorem are not absolute necessity for f'(x) to be zero at some point in the concerned interval. iii) The function f defined in [-1,1] as follows: f(x)= . Here, f(x) is continuous on [-1,1], f(-1)=f(1), but f'(0) does not exists i.e. , f(x) is not differentiable on (-1,1). Also, f'(x) vanishes nowhere in (-1,1)-{0}. Failure of Rolle’s Theorem can be explained by the fact that is not derivable in -1<x<1, though other conditions are satisfied. iv) The function f defined in [-1,1] as follows: f(x)=x3+5x-4. f(x) is continuous on [-1,1], f(x) is derivable on (-1,1), but f(-1)≠f(1). Also, f'(x) =3x2+5 vanishes nowhere in (-1,1). Failure of Rolle’s Theorem can be explained by the fact that , though other conditions are satisfied.
  • 5.
    v) The functionf defined in [-1,1] as follows: f(x)=x2, x=-1 = 5x , -1<x<1 = x2, x=1 Here, f(x) is continuous on (-1,1) and discontinuous at the points x=-1 and x=1. i.e., f(x) is not continuous on [-1,1], f(x) is derivable on (-1,1), and f(-1)=f(1). f'(x) vanishes nowhere in (-1,1). Failure of Rolle’s Theorem can be explained by the fact that f(x) is not continuous in -1 1, though other conditions are satisfied. Examples (iii), (iv)& (v) show that if we drop any of the conditions in Rolle’s Theorem then Rolle’s Theorem fails to be true. Important Note: If f(x) satisfies all the conditions of Rolle’s Theorem in [a,b] then the conclusion that f (c)=0 where a<c<b is assured, but if any of the conditions are violated then Rolle’s Theorem will not be necessarily true; it may still be true but the truth is not ensured.
  • 6.
    Lagrange’s Mean ValueTheorem (First Mean Value Theorem of Differential Calculus) Statement Let a function f:[a,b]R be such that i) f is continuous on [a,b], ii) f is differentiable on (a,b). Then there exists at least one point c(a,b) such that Another Form of Lagrange’s MVT: If in the statement of the theorem, b is replaced by a+h, (h>0), then the number between a and b may be written as a+θh, where 0<θ<1. Then the theorem takes the following form: Let f:[a,a+h]R be such that iii) f is continuous on [a,a+h], iv) f is differentiable on (a,a+h). Then there exists a real number θ lying between 0 and 1 such that
  • 7.
    or, f(a+h)=f(a)+hf′(a+θh). ** MeanValue Theorem relates the mean rate of change to instantaneous rate of the function. Somewhere inside a chord, the tangent to f will be parallel to the chord. The accurate statement of this common-sense observation is the Mean Value Theorem
  • 8.
    For any functionthat is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to thetangent at c. Geometrical Interpretation: If the function y=f(x) has a graph which is a continuous curve on [a,b], and the curve has a tangent at every point of (a,b),then there exists at least one point c in (a,b) such that the tangent to the curve at (c,f(c)) is parallel to the line segment joining the points (a,f(a)) and (b,f(b)). Remarks: i) The fraction measures the mean (or average) rate of increase of the function in the interval [a,b] of length b-a. Hence the theorem expresses the fact that, under the conditions stated, the mean rate of increase in any interval [a,b] is equal to the actual rate of increase at
  • 9.
    some point cwithin the interval (a,b). For instance, the mean velocity of a moving car in any interval of time is equal to the actual velocity at some instant within the interval. This justifies the name Mean Value Theorem. ii) Mean Value Theorem proposes that any differentiable function defined over an interval has a mean value at which a tangent line is parallel to the line joining the end points of the function’s graph on that interval. iii) Rolle’s Theorem is a particular case of Lagrange’s Mean Value Theorem. If f(a)=f(b) holds in addition to the two conditions of Mean Value Theorem then f(b)-f(a)=0 and consequently f (c)=0. Important deductions: i) Let f:[a,b]R be such that *f is continuous on [a,b], *f is differentiable on (a,b) And * f'(x)=0 for all x(a,b). Then f(x) is a constant function on [a,b]. ii) Let f:[a,b]R be such that *f is continuous on [a,b], *f is differentiable on (a,b) and * f'(x) 0 for all x(a,b). Then f(x) is a monotone increasing function on [a,b].
  • 10.
    iii) Let f:[a,b] R be such that *f is continuous on [a,b], *f is differentiable on (a,b) and * f'(x) 0 for all x(a,b). Then f(x) is a monotone decreasing function on [a,b]. Cauchy’s Mean Value Theorem: (Second Mean Value Theorem of Differential Calculus) Statement Let the functions f:[a,b]R and g:[a,b]R be such that i) both f,g are continuous on [a,b], ii) both f,g are differentiable on (a,b), and iii) f'(x)≠0 for x(a,b). Then there exists at least one point (a,b) such that .
  • 11.
    Second Statement: Let the functions f:[a,a+h]R and g:[a,a+h]R be such that iv) both f,g are continuous on [a,a+h], v) both f,g are differentiable on (a,a+h), and vi) f'(x)≠0 for x(a,a+h). Then there exists at least one real number θ lying between 0 and 1 such that . Geometric Interpretation of Cauchy’s MVT: First Interpretation: The functions f and g can considered as determining a curve in the plane by means of parametric equations x=f(t),y=g(t),where atb. Cauchy’s MVT concludes that  a point (f(),g())of the curve for some t= in (a,b) such that the slope of the line segment joining the end points (f(a),g(a)) and (f(b),g(b)) of the curve is equal to the slope of the tangent to the curve at t=.
  • 12.
    Second Interpretation: Wemay write Hence, the ratio of the mean rates of increase of two functions in an interval [a,b] is equal to the ratio of the actual rates of increases of the functions at some point  within the interval (a,b). REMARKS: i) Lagrange’s MVT can be deduced from Cauchy’s MVT by taking f(x)=x, x(a,b). ii) Rolle’s Theorem can be obtained from Cauchy’s MVT by letting f(x)=x and g(b)=g(a). [We have used Rolle’s Theorem to prove Cauchy’s MVT.]
  • 13.
    iii) Both fand g satisfy the conditions of Lagrange’s MVT. Consequently there exist points c and d in (a,b) such that and c and d are different points in (a,b) in general, and therefore a single point  in (a,b) may not be found to satisfy the conclusion of Cauchy’s MVT. Some problems: Q1. Show that the equation 4x5+x3+7x-1=0 has exactly one real root. Q2. A function f is differentiable on [0,2] and f(0)=0,f(1)=1,f(2)=1. Prove that f'(c)=0 for some c in (0,2). Q3. Prove that the equation (x-1)3+(x-2)3+(x-3)3+(x- 4)3=0 has only one real root. Q4. Prove that between any two real roots of excosx+1=0 there is at least one real root of exsinx+1=0. Q5. Verify Lagrange’s MVT for the function f(x) =x(x-1)(x-2) in the interval [0, ]. Q6. Show that if 0<u<v deduce that < < .
  • 14.
    Q7. If in the Cauchy’s MVT we take f(x)= and g(x)= , then prove that  is the arithmetic mean between a and b. (Refer to the statement of Cauchy’s MVT). Q8. If f(x)= (x-a)m(x-b)n where m and n are positive integers, show that c in Rolle’s theorem divides the segment a x b in the ratio m:n. Q9. Apply MVT to prove that lies between 10 and 10.05. Q10. Show that , where 0< .