Rolle's Mean Value TheoremRoll's theorem is a special case of the mean value theorem. Rolle's theorem states that if a f be a real valued function defined on the closed interval [a, b] where (a?x ?b) and differentiable in open interval ]a,b[where (a < x< b), and f (a) = f(b) then there is at least one value c in ]a,b[for which Proof of Rolle's mean value theorem If the function f is continuous in [a,b] and differentiable in open interval ]a,b[, with f(a) = f (b)=0 then there exist c in (a,b) where f' (c) = 0. Proof Let's suppose the two cases that could happen: Case 1: f(x) = 0 for all x in [a,b]. In the above case, the value that exists between a and b can serve as the c (mentioned in the equation), as the function is constant on [a,b], and as we know, the derivatives of constant functions are always zero. Case 2: f(x) = ? 0 for x in open interval ]a,b[ We know by the extreme value theorem that f accomplishes both its absolute minimum and maximum values somewhere on [a,b]. As we discussed above, f(a) = f (b)=0, and that in this case, f(x) is not equal to zero for some x in open interval ]a,b[. Therefore, f will have either positive absolute maximum value at some C_{max} in closed interval [a,b] or a negative absolute minimum values in closed interval some C_{min} in ]a,b[ or both. Take c to be either C_{max} or C_{min}, depending on which you have In the open interval ]a,b[ contains c, and either;
It means f has a local extremum at c. As f is also differentiable at c, Fermat's theorem applies and concludes that f' (c) = 0 Let's understand this concept with the help of an example Example 1: Verify Rolle's theorem for the function y = x^{2} + 3, a =  2 and b = 2 Solution: We know the statement of the Rolle's theorem, the function y = = x^{2} + 3 is continuous in [2 ,2] and differentiable in (2,2) Given; F(x) = x^{2 }+ 3 F (2) = (2)^{2} + 3 = 4 + 3 = 7 F (2) = (2)^{2} + 3 = 4 + 3 = 7 Thus, F (2) = F (2) Therefore, the value of f (x) at 2 and 2 coincide Now, F' (x) = 2x At c = 0, F'(c) = 2 (0), where c=0 ∈ (2,2) Example 2: For all seconddegree polynomials with y = mx^{2} + nx + k, it is seen that the Rolle's point is at c = 0. Also, the value of k is zero. Then find the value of n? Solution: Given; K = 0 We have to rewrite the function to get y = mx^{2} + nx Now, differentiating the function yields Y' = 2mx + n = 0 Equating it to zero we get the Rolle's point which is also zero 2m(0) + = 0 n= 0 Geometric InterpretationThe geometrical meaning of Rolle's mean value theorem states that the curve y = f (x) is continuous between x = a and x = b. At every point of time, within the interval, it is possible to make a tangent and ordinates corresponding to the abscissa and are equal then exists at least one tangent to the curve which is parallel to the xaxis. Algebraically, Rolle's theorem states that if f (x) is showing a polynomial function in x and the two roots of the equation f (x) = 0 and x = a and x = b, then there exists at least one root of the equation f (x) = 0 lying between these values. Questions based on Rolle's TheoremExamples 1 Verify Rolle's theorem for the function y = x^{2 }+ 5, a =  3 and b = 3 Solution: We know the statement of the Rolle's theorem, the function y = = x^{2 }+ 5 is continuous in [3 ,3] and differentiable in (3,3) Given; F(x) = x^{2 }+ 5 F (3) = (3)^{2} + 5 = 9 + 5 = 14 F (3) = (3)^{2} + 5 = 9 + 5 = 14 Thus, F (3) = F (3) Therefore, the value of f (x) at 3 and 3 coincide Now, F' (x) = 2x At c = 0, F'(c) = 3 (0), where c=0 ∈ (3,3) Example 2: For all seconddegree polynomials with y = ix^{2} + jx + k, it is seen that the Rolle's point is at c = 0. Also, the value of k is zero. Then find the value of j? Solution: Given; K = 0 We have to rewrite the function to get y = ix^{2} + jx Now, differentiating the function yields Y' = 2ix + j = 0 Equating it to zero we get the Rolle's point which is also zero 2i(0) + j = 0 J = 0
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