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Mathematics | Lagrange’s Mean Value Theorem

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


 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., 


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(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 




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 

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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