{"id":5585,"date":"2021-09-22T01:41:52","date_gmt":"2021-09-21T20:11:52","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5585"},"modified":"2022-01-16T17:03:59","modified_gmt":"2022-01-16T11:33:59","slug":"rolles-theorem","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/rolles-theorem\/","title":{"rendered":"Rolle’s Theorem – Statement and Examples"},"content":{"rendered":"
Here you will learn statement of rolle’s theorem, it’s geometrical and algebraic interpretation with examples.<\/p>\n
Let’s begin –<\/p>\n
\nStatement<\/strong> : Let f be a function that satisfies the following three conditions:<\/p>\n
(a) f is continous on the closed interval [a, b].<\/p>\n
(b) f is differentiable on the open interval (a, b)<\/p>\n
(c) 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
Geometrical Interpretation :<\/strong><\/h4>\n
\nGeometrically, the theorem says that somewhere between A and B the curve has atleast one tangent parallel to x-axis.<\/p>\n<\/blockquote>\n
Algebraic Interpretation :<\/strong><\/h4>\n
\nIf 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<\/blockquote>\n
Remark – <\/strong>On this theorem generally two types of problems are formulated.<\/p>\n
(a) To check the applicability of rolle’s theorem to a given function on a given interval.<\/p>\n
(b) To verify rolle’s theorem for a given function on a given interval. In both types of problems we first check whether f(x) satisfies conditions of theorem or not. The following results are very helpful in doing so.<\/p>\n
\n
- A polynomial function is everywhere continuous and differentiable.<\/li>\n
- The exponential function, sine and cosine functions are everywhere continuous and differentiable.<\/li>\n
- Logarithmic function is continuous and differentiable in its domain.<\/li>\n
- | x | is not differentiable at x = 0<\/li>\n<\/ol>\n
Example<\/span><\/strong> : Verify Rolle’s theorem for the function f(x) = \\(x^2\\) – 5x + 6 on the interval [2, 3].<\/p>\n
Solution<\/strong><\/span> : Since a polynomial function is everywhere differentiable and so continuous also.<\/p>\n
Therefore, f(x) is continuous on [2, 3] and differentiable on (2, 3).<\/p>\n
Also, f(2) = \\(2^2\\) – 5 \\(\\times\\) 2 + 6 = 0 and f(3) = \\(3^2\\) – 5 \\(\\times\\) 3 + 6 = 0<\/p>\n
\\(\\therefore\\) f(2) = f(3)<\/p>\n
Thus, all the conditions of rolle’s theorem are satisfied. Now, we have to show that there exists some c \\(\\in\\) (2, 3) such that f'(c) = 0.<\/p>\n
for this we proceed as follows,<\/p>\n
We have, <\/p>\n
f(x) = \\(x^2\\) – 5x + 6 \\(\\implies\\) f'(x) = 2x – 5<\/p>\n
\\(\\therefore\\) f'(x) = 0 \\(\\implies\\) 2x – 5 = 0 \\(\\implies\\) x = 2.5<\/p>\n
Thus, c = 2.5 \\(\\in\\) (2, 3) such that f'(c) = 0. Hence, Rolle’s theorem is verified.<\/p>\n
\nRelated Questions<\/h3>\n