2. ROLLE’S THEOREM
Statement:-
If f(x) is a function of the variable x such that :-
f(x) is continuous in the closed interval [a , b].
f(x) is differentiable for every point in the open interval (a ,
b).
f(a)= f(b), then there is at least one point c in the open
interval (a , b) for which f’(c) = 0
3. ROLLE’S THEOREM GEOMETRICAL INTERPRETATION
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From Rolle's Theorem, you can see that if a function f is continuous on [a, b]
and differentiable on (a, b), and if f(a) = f(b), then there must be at least one
x-value between a and b at which the graph of f has a horizontal tangent, as
shown in Figure 1.
Figure 1
4. ROLLE’S THEOREM GEOMETRICAL INTERPRETATION
4
If the differentiability requirement is dropped from Rolle's
Theorem, f will still have a critical number in (a, b), but it may not
yield a horizontal tangent. Such a case is shown in Figure 2
f is continuous on [a , b]
Figure 2
5. ROLLE’S THEOREM SOME COUNTERCAMPLES
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1. Consider f(x) = ( {x} is the fractional part function ) on [0,1].The derivative of the function on
(0,1) = 1 everywhere . In this case ,Rolle’s Theorem fails because the function f(x) has discontinuity
at x = 1 (i-e, it is not continuous everywhere on [0,1]
6. ROLLE’S THEOREM SOME COUNTERCAMPLES
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2. The linear function f(x) = x is continuous on [o,1] and differentiable on (0,1) .But, Rolle’s
Theorem fails for the function as f(0) = 0 & f(1) = 1 [cause – f(0) ≠ f(1)]
7. EXAMPLE
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Let f(x) = 𝑥2 + 2𝑥
Find the applicability of Rolle’s theorem on [ -2 , 0 ]. Hence, find all values of c in [ 2, 0 ]
Solution :-
i) f(x) is continuous in [ -2 , 0 ] as a quadratic function.
ii) It is differentiable everywhere over the ( -2 , 0 )
iii) 𝑓 2 = −2 2 + 2. −2 = 0
𝑓 0 = 02 + 2.0 = 0
so,𝑓 −2 = 𝑓 0
So , we can apply Rolle’s Theorem
To find the point c ,let us find
𝑓′
𝑥 = 2𝑥 + 2
𝑓′
𝑐 = 0
2𝑐 + 2 = 0
2𝑐 = −2
𝑐 = −1
𝑓′
𝑐 = 0 𝑓𝑜𝑟 𝑐 = −1