Given as f(x) = x + (1/x) on [1,3]
f(x) has unique value for all x ∈ (1,3)
So, f(x) is continuous in [1,3]
f(x) = x + (1/x)
Differentiate with respect to x
f'(x) exists for the all values except 0
So, f(x) is differentiate in (1,3)
Therefore, both the necessary condition of lagrange'* mean value theorem is satisfied. So, there exist a point c ∈ (1,3)
f'(c) = (f(3) - f(1))/(3 - 1)
f'(c) = (f(3) - f(1))/2
f'(x) = x + (1/x)
Differentiate with respect to x
f'(x) = (x2 - 1)/x2
For the f'(c), put the value of x = c in f'(x)
f'(c) = (c2 - 1)/c2
For the f'(3), put the value of x = 3 in f'(x)
f(3) = 3 + (1/3)
f(3) = (9 + 1)/3
f(3) = 10/3
For the f'(1), put the value of x = 1 in f'(x)
Thus, lagrange'* mean value theorem is verified.
