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Lagrange’s Mean Value Theorem

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Suppose 

[Tex]f:[a,b]\rightarrow R   [/Tex]
 

be a function satisfying these conditions:



 

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}   [/Tex]
 


 We can visualize Lagrange’s Theorem by the following figure



 



 



 



 

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., 


[Tex]f'(c)=\frac{f(b)-f(a)}{b-a}  [/Tex]
 


Example: Verify mean value theorem for f(x) = x2 in interval [2,4]. 

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. 
 


[Tex]{f}'(x)=2x   [/Tex]

f(2) = 4 


and f(4) = 16 

[Tex]\frac{f(b)-f(a)}{b-a} = \frac{16-4}{4-2}=6   [/Tex]
 
Mean value theorem states that there is a point c Є (2, 4) such that 


[Tex]{f}'(c)=6   [/Tex]
 

But 


[Tex]{f}'(x)=2x   [/Tex]
 

which implies c = 3. Thus at c = 3 Є (2, 4), we have 


[Tex]{f}'(c)= 6   [/Tex]


This article has been contributed by Saurabh Sharma. 
  
If you would like to contribute, please email us your interest at review-team@geeksforgeeks.org 


Last Updated : 14 Mar, 2024
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