Given as f(x) = 1/(4x - 1), on [1,4]
Here, 4x - 1 > 0
f'(x) has the unique values for all x except 1/4
So, f(x) is continuous in [1,4]
f(x) = 1/(4x - 1)
Differentiate with respect to x
f'(x) = (-1)(4x - 1)-2(4)
f'(x) = - 4/(4x - 1)2
Where, 4x - 1 > 0
f'(x) has the unique values for all x except 1/4
So, f(x) is differentiable in (1,4)
Therefore both the necessary conditions of lagrange's mean value theorem is satisfied. So, there exist a point c ∈ (1,4)
Differentiate with respect to x
f'(x) = -4/(4x - 1)2
For the f'(c), put the value of x = c in f'(x)
f'(c) = -4/(4c - 1)2
For the f'(4), put the value of x = 4 in f'(x)
For the f'(1), put the value of x = 1 in f'(x)
Thus, lagrange's mean value theorem is verified.