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.
 
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