Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that
f(b) − f(a) = f′(c)(b − a)
\(\Rightarrow \text{f'(c)} = \frac{f(b) - f(a)}{b - a}\)
This theorem is also known as First Mean Value Theorem.
\(\Rightarrow\) f(x) has unique values for all x except \(\frac{1}{4}\)
∴ f(x) is continuous in [1, 4]
f(x) = \(\frac{1}{4x - 1}\)
Differentiating with respect to x:
\(\Rightarrow\) f'(x) has unique values for all x except \(\frac{1}{4}\)
∴ f(x) is differentiable in (1, 4)
So both the necessary conditions of Lagrange’s mean value theorem is satisfied.
Therefore, there exists a point c∈(1, 4) such that:
On differentiating with respect to x:
f'(x) = \(-\frac{4}{(4x - 1)^2}\)
For f’(c), put the value of x = c in f’(x):
f'(c) = \(-\frac{4}{(4c - 1)^2}\)
For f(4), put the value of x = 4 in f(x):
For f(1), put the value of x = 1 in f(x):
Hence, Lagrange’s mean value theorem is verified.
