Question

Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each case find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem: f(x) = x² + x – 1 on [o, 4]

Answer 1

f(x) = x² + x – 1 on (0, 4) Every polynomial function is continuous everywhere on (- ∞, ∞) and differentiable for all arguments. Here, f (x) is a polynomial function. So, it is continuous in 10,4] and differentiable on (0,4). So, both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point CE (0, 4) such that:

Now, f(x) = x² + x – 1

Differentiating both sides w.r.t.’x’,

we get

f(x) = 2c + 1

For f’(4), put the value of x = 4 in f(x):

f'(c) = 2c + 1

For f(4), put the value of x = 4 in f(x):

f(4) = 4² + 4 – 1 = 16 + 4 – 1 = 19

For) (0), put the value of x = 0 in f(x):

f(0) = 0² + 0 – 1 = 0 + 0 – 1 = – 1

Now, f’(c) = f(4)−f(0)/4

⇒ 2c + 1 = 19−(−1)/4 = 20/4

⇒ 2c + 1 = 5 ⇒ 2c = 5 – 1 = 4

⇒ c = 4/2 = 2 ∈ (0,4)

Hence, Lagrange’s mean value theorem is verified.