Rolle's Mean Value Theorem in Calculus
Rolle’s Mean Value
theorems
B.Sc. I Year
Mr. Shrimangale G.W.
Rolle’s mean value theorem
Theorem :
- If a function f is ,
1)
Continuous in a closed interval [a,b]
2) Derivable in the open interval (a,b) and
3) f(a) = f(b), then there exists at least one value
c
ϵ
(a,b) such that f’(c) = 0.
Proof :-
By virtue of continuity of the function in the
closed interval [a,b] it has a greatest value M and
a least value m in the interval, so that there two
numbers c and d such that
f(c) = M and f(d) = m
Now either M = m ………………………..(i)
or M ≠ m ………………………..(ii)
When the greatest value coinsides with the least
value as in case (i), the function reduces to a
constant so that the derivative ia equal to 0 for
every value of x and therefore the theorem is
true in this case.
When M and m are unequal,as in case
(ii),at least one of them must be
different from the equal values f(a)
and f(b).Let M=f(c) be different from
them. The number c being different
from a and b,lies within the interval
[a,b] and as such belongs to the open
interval (a,b),in particular , derivative
at c, so that
when h → 0
exists and the same when h→0 through
positive value.Also f(c) is the greatest value
of the function , we have
f(c+h) ≤ f(c) whatever positive or negative
value h has.Thus
(iii)……
(iv)……
Let h→0 through positive value. From
(iii),we get f’(c) ≤ 0………….(v)
Let h→0 through negative value. From
(iv),we get f’(c)≥0…………(vi)
the relation (v) and (vi)will both true if and
only if
f’(c) = 0.
Thus the same conclusion would be similar
reached if it is the least value which differ
from f(a) and f(b).
Hence the theorem.
Example-1
Ex.
Verify the Rolle’s theorem for
Solution :-
Here from the given function it is
clear that f(1) = 1 = f(-1).
The conditions of the theorem being satisfied,
the derivative f’(x) must vanish for at least one
of x
ϵ
(-1,1).
Directly we see that the derivative vanishes for
x=0 which belongs to (-1,1).Hence the
verification.
Example-2
Example :-
Verify the Rolle’s theorem for
Solution :-
Let
So that f(-3) = 0 = f(0).
Also the function is derivable in the interval
[-3,3].we have
Now f’(x)=0
The equation is satisfied by
x=-2,3.Of these two values of x for
which f’(x) is zero,-2 belongs to the
open interval (-3,0) under
consideration. Hence the
verification.
Problems for home work :-
Examples :-
Verify the Rolle’s theorem for the
function:
(i)
(ii)
(iii)
Thank You
Rolle's Mean Value Theorem states that if a function is continuous in a closed interval, differentiable in the open interval, and the function values at the endpoints of the interval are equal, then there exists at least one point where the derivative of the function is zero. This theorem is verified through examples and proofs, showcasing its significance in calculus.
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Presentation Transcript
Rolles Mean Value theorems B.Sc. I Year Mr. Shrimangale G.W.
Rolles mean value theorem Theorem :- If a function f is , 1) Continuous in a closed interval [a,b] 2) Derivable in the open interval (a,b) and 3) f(a) = f(b), then there exists at least one value c (a,b) such that f (c) = 0.
Proof :- By virtue of continuity of the function in the closed interval [a,b] it has a greatest value M and a least value m in the interval, so that there two numbers c and d such that f(c) = M and f(d) = m Now either M = m ..(i) or M m ..(ii) When the greatest value coinsides with the least value as in case (i), the function reduces to a constant so that the derivative ia equal to 0 for every value of x and therefore the theorem is true in this case.
When M and m are unequal,as in case (ii),at least one of them must be different from the equal values f(a) and f(b).Let M=f(c) be different from them. The number c being different from a and b,lies within the interval [a,b] and as such belongs to the open interval (a,b),in particular , derivative at c, so that when h 0
exists and the same when h0 through positive value.Also f(c) is the greatest value of the function , we have f(c+h) f(c) whatever positive or negative value h has.Thus (iii) (iv)
Let h0 through positive value. From (iii),we get f (c) 0 .(v) Let h 0 through negative value. From (iv),we get f (c) 0 (vi) the relation (v) and (vi)will both true if and only if f (c) = 0. Thus the same conclusion would be similar reached if it is the least value which differ from f(a) and f(b). Hence the theorem.
Example-1 Ex. Verify the Rolle s theorem for Solution :- Here from the given function it is clear that f(1) = 1 = f(-1). The conditions of the theorem being satisfied, the derivative f (x) must vanish for at least one of x (-1,1). Directly we see that the derivative vanishes for x=0 which belongs to (-1,1).Hence the verification.
Example-2 Example :- Verify the Rolle s theorem for Solution :- Let So that f(-3) = 0 = f(0). Also the function is derivable in the interval [-3,3].we have Now f (x)=0
The equation is satisfied by x=-2,3.Of these two values of x for which f (x) is zero,-2 belongs to the open interval (-3,0) under consideration. Hence the verification.
Problems for home work :- Examples :- Verify the Rolle s theorem for the function: (i) (ii) (iii)
