Verify Rolle’s theorem for each of the following functions on the indicated intervals :
f(x) = x2 + 5x + 6 on [ – 3, – 2]
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):
⇒
⇒
⇒
⇒ f’(x) = 2x + 5
We have f’(c) = 0 cϵ( – 3, – 2), from the definition given above.
⇒ f’(c) = 0
⇒ 2c + 5 = 0
⇒ 2c = – 5
⇒
⇒ C = – 2.5ϵ( – 3, – 2)
∴ Rolle’s theorem is verified.