Mathematics | Lagrange’s Mean Value Theorem

Suppose f:[a,b]\rightarrow R be a function satisfying three 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

f3

 

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

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up

Article Tags :

2


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.