{"id":5583,"date":"2021-09-22T00:24:36","date_gmt":"2021-09-21T18:54:36","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5583"},"modified":"2021-11-14T17:58:52","modified_gmt":"2021-11-14T12:28:52","slug":"mean-value-theorems-class-12","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/mean-value-theorems-class-12\/","title":{"rendered":"Mean Value Theorems Class 12"},"content":{"rendered":"
Here you will learn mean value theorems i.e rolle’s theorem, lagrange’s theorem and extreme value theorem.<\/p>\n
Let’s begin –<\/p>\n
Let f be a real valued function defined on the closed interval [a, b] such that<\/p>\n
\n(i) it is continuous on the closed interval [a, b],<\/p>\n
(ii) it is differentiable on the open interval (a, b)<\/p>\n
(iii) f(a) = f(b)<\/p>\n
Then, there exist a real number c \\(\\in\\) (a, b) such that f'(c) = 0<\/p>\n<\/blockquote>\n
Note<\/strong> : If f is differentiable function then between any two consecutive roots of f(x) = 0, there is atleast one root of the equation f'(x) = 0<\/p>\n
(b) Lagrange’s Mean Value Theorem (LMVT)<\/a><\/strong><\/h4>\n
Let f be a function that satisfies the following conditions :<\/p>\n
\n(i) f is continuous in [a, b]<\/p>\n
(ii) f is differentiable in (a, b).<\/p>\n
Then there is a number c in (a, b) such that f'(c) = \\(f(b) – f(a)\\over b – a\\)<\/p>\n<\/blockquote>\n
Extreme Value Theorem\u00a0<\/h3>\n
If f is continuous on [a, b] then f takes on,\u00a0 a least value m and a greatest value M on this interval.<\/p>\n
Note<\/strong> : Continuity throught the interval [a, b] is essential for the validity of this theorem.<\/p>\n
(a) If a continuous function y = f(x) is increasing in the closed interval [a, b] , then f(a) is the least value and f(b) is the greatest value of f(x) in [a, b]<\/p>\n
(b) If a continuous function y = f(x) is decreasing in the closed interval [a, b] , then f(b) is the least value and f(a) is the greatest value of f(x) in [a, b]<\/p>\n
(c) If a continuous function y = f(x) is increasing\/decreasing in the (a, b) , then no greatest and least value exist.<\/p>\n\n\n