Question

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

f(x) = x² + 5x + 6 on [– 3, – 2]

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:

⇒ f(x) = x² + 5x + 6 on [– 3, – 2]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremums:

⇒ f(– 3) = (– 3)² + 5(– 3) + 6

⇒ f(– 3) = 9 – 15 + 6

⇒ f(– 3) = 0

⇒ f(– 2) = (– 2)² + 5(– 2) + 6

⇒ f(– 2) = 4 – 10 + 6

⇒ f(– 2) = 0

∴ f(– 3) = f(– 2), Rolle’s theorem applicable for function ‘f’ on [– 3, – 2].

Let’s find the derivative of f(x):

We have f’(c) = 0 c∈(– 3, – 2), from the definition given above.

⇒ f’(c) = 0

⇒ 2c + 5 = 0

⇒ 2c = – 5

⇒ c = \(-\frac{5}{2}\)

⇒ C = – 2.5∈(– 3, – 2)

∴ Rolle’s theorem is verified.