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Mathematics | Lagrange’s Mean Value Theorem

  • Difficulty Level : Medium
  • Last Updated : 16 Jul, 2021

Suppose 

f:[a,b]\rightarrow R
 

be a function satisfying these conditions:

 

1) f(x) is continuous in the closed interval a ≤ x ≤ 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:

f'(c)=\frac{f(b)-f(a)}{b-a}
 

 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., 

f'(c)=\frac{f(b)-f(a)}{b-a}
 

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. 
 

{f}'(x)=2x

f(2) = 4 

and f(4) = 16 

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

{f}'(c)=6
 



But 

{f}'(x)=2x
 

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

{f}'(c)= 6

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

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