{"id":5587,"date":"2021-09-22T02:26:53","date_gmt":"2021-09-21T20:56:53","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5587"},"modified":"2021-11-14T17:20:13","modified_gmt":"2021-11-14T11:50:13","slug":"lagranges-mean-value-theorem","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/lagranges-mean-value-theorem\/","title":{"rendered":"Lagrange’s Mean Value Theorem"},"content":{"rendered":"
Here you will learn lagrange’s mean value theorem statement, its geometrical and physical interpretation with examples.<\/p>\n
Let’s begin – <\/p>\n
\nStatement<\/strong> : Let f be a function that satisfies the following conditions :<\/p>\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
Geometrical Interpretation :<\/strong><\/h4>\n
\nGeometrically, the lagrange’s mean value theorem says that somewhere between A and B the curve has atleast on tangent parallel to chord AB.<\/p>\n<\/blockquote>\n
Physical Interpretation :<\/strong><\/h4>\n
\nIf we think of the number (f(b) – f(a))\/(b – a) as the average change in f over [a, b] and f'(c) as an instantaneous change, then the mean value theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : find c of the Lagranges mean value theorem for the function f(x) = \\(3x^2\\) + 5x + 7 in the interval [1, 3].<\/p>\n
Solution<\/strong><\/span> : Given f(x) = \\(3x^2\\) + 5x + 7<\/p>\n
\\(\\implies\\) f(1) = 3 + 5 + 7 = 15 and f(3) = 27 + 15 + 7 = 49<\/p>\n
Now, Differentiating f(x) with respect to x,<\/p>\n
\\(\\implies\\) f'(x) = 6x + 5<\/p>\n
Here a = 1, b = 3<\/p>\n
Now from lagrange’s mean value theorem <\/p>\n
f'(c) = \\(f(b) – f(a)\\over b – a\\) \\(\\implies\\) 6c + 5 = \\(49 – 15\\over 2\\) = 17<\/p>\n
\\(\\implies\\) c = 2<\/p>\n
Example<\/strong><\/span> : If f(x) is continuous and differentiable over [-2, 5] and -4 < f'(x) < 3 for all x in (-2, 5), then the greatest possible value of f(5) – f(-2) is –<\/p>\n
Solution<\/strong><\/span> : Applying Lagranges mean value theorem (LMVT),<\/p>\n
f'(x) = \\(f(5) – f(-2)\\over 5 -(-2)\\) for some x in (-2, 5) <\/p>\n
Now, -4 \\(\\le\\) \\(f(5) – f(-2)\\over 7\\) \\(\\le\\) 3<\/p>\n
-28 \\(\\le\\) f(5) – f(-2) \\(\\le\\) 21<\/p>\n
\\(\\therefore\\) Greatest possible value of f(5) – f(-2) is 21.<\/p>\n\n\n