Verify Rolle’s theorem for each of the following functions on the indicated intervals : f(x) = x/2 - sin πx/6 on [-1, 0]

Question

Verify Rolle’s theorem for each of the following functions on the indicated intervals :

\(\text{f(x)}=\frac{\text{x}}{2}-\text{sin}\frac{\text{πx}}{6}\) on [-1, 0]

Answer 1

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a, b].

b) The function ‘f’ needs differentiable on the open interval (a, b).

c) f(a) = f(b)

Then there exists at least one c in the open interval (a, b) such that f’(c) = 0.

Given function is:

\(\Rightarrow f(x) = \frac{x}{2}-sin\big(\frac{\pi x}{6}\big)\) on [-1, 0]

We know that sine function is continuous and differentiable over R.

Let’s check the values of ‘f’ at an extremum

We have got f(– 1) = f(0). So, there exists a c∈(– 1, 0) such that f’(c) = 0.

Let’s find the derivative of the function ‘f’

Cosine is positive between \(-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2},\) for our convenience we take the interval to be \(-\frac{\pi}{2}\leq \theta \leq 0,\) since the values of the cosine repeats.

We know that \(\frac{3}{\pi}\) value is nearly equal to 1. So, the value of the c nearly equal to 0.

So, we can clearly say that c∈(– 1, 0).

∴ Rolle’s theorem is verified.