Lagrange’s Mean Value Theorem
SupposeÂ
[Tex]f:[a,b]\rightarrow RÂ Â Â [/Tex]
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be a function satisfying these conditions:
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1) f(x) is continuous in the closed interval [a, b]
2) f(x) is differentiable in the open interval a < x < b
Then according to Lagrange’s Theorem, there exists at least one point ‘c’ in the open interval (a, b) such that:
[Tex]f'(c)=\frac{f(b)-f(a)}{b-a}Â Â Â [/Tex]
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 We can visualize Lagrange’s Theorem by the following figure
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In simple words, Lagrange’s theorem says that if there is a path between two points A(a, f(a)) and B(b, f(a)) in a 2-D plain then there will be at least one point ‘c’ on the path such that the slope of the tangent at point ‘c’, i.e., (f ‘ (c)) is equal to the average slope of the path, i.e.,Â
[Tex]f'(c)=\frac{f(b)-f(a)}{b-a}Â Â [/Tex]
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Example: Verify mean value theorem for f(x) = x2 in interval [2,4].Â
Solution: First check if the function is continuous in the given closed interval, the answer is Yes. Then check for differentiability in the open interval (2,4), Yes it is differentiable.Â
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[Tex]{f}'(x)=2x   [/Tex]
f(2) = 4Â
and f(4) = 16Â
[Tex]\frac{f(b)-f(a)}{b-a} = \frac{16-4}{4-2}=6Â Â Â [/Tex]
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Mean value theorem states that there is a point c Є (2, 4) such thatÂ
[Tex]{f}'(c)=6Â Â Â [/Tex]
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ButÂ
[Tex]{f}'(x)=2x   [/Tex]
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which implies c = 3. Thus at c = 3 Є (2, 4), we haveÂ
[Tex]{f}'(c)= 6Â Â Â [/Tex]
This article has been contributed by Saurabh Sharma.Â
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If you would like to contribute, please email us your interest at review-team@geeksforgeeks.orgÂ
Last Updated :
14 Mar, 2024
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