Given as f(x) = x2 + x – 1 on [0, 4]
The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Since, f(x) is a polynomial function. Therefore it is continuous in [0, 4] and differentiable in (0, 4). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied. So, there exist a point c ∈ (0, 4) such that:
f'(c) = (f(4) - f(0))/(4 - 0)
f'(c) = (f(4) - f(0))/4
f (x) = x2 + x – 1
Differentiate with respect to x:
f’(x) = 2x + 1
For the f’(c), put the value of x = c in f’(x):
f’(c) = 2c + 1
For the f(4), put the value of x = 4 in f(x):
f (4) = (4)2 + 4 – 1
= 16 + 4 – 1
= 19
For the f(0), put the value of x = 0 in f(x):
f (0) = (0)2 + 0 – 1
= 0 + 0 – 1
= – 1
Thus, lagrange's mean value theorem is verified.