Verify Rolle’s theorem for each of the following functions on the indicated intervals :
f(x) = x2 – 8x + 12 on [2, 6]
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 – 8x + 12 on [2,6]
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(2) = 22 – 8(2) + 12
⇒ f(2) = 4 – 16 + 12
⇒ f(2) = 0
⇒ f(6) = 62 – 8(6) + 12
⇒ f(6) = 36 – 48 + 12
⇒ f(6) = 0
∴ f(2) = f(6), Rolle’s theorem applicable for function ‘f’ on [2,6].
Let’s find the derivative of f(x):
⇒
⇒
⇒
⇒ f’(x) = 2x – 8
We have f’(c) = 0 cϵ(2,6), from the definition given above.
⇒ f’(c) = 0
⇒ 2c – 8 = 0
⇒ 2c = 8
⇒
⇒ C = 4ϵ(2,6)
∴ Rolle’s theorem is verified.