Skip to content
Related Articles

Related Articles

Improve Article
Mathematics | Lagrange’s Mean Value Theorem
  • Difficulty Level : Medium
  • Last Updated : 01 Oct, 2020

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 theorm 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 theorm 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 contribute@geeksforgeeks.org 

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

Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :