Verify Rolle’s theorem for each of the following functions on the indicated intervals :
f(x) = x(x – 1)2 on [0, 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) = x(x – 1)2 on [0,1]
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(0) = 0(0 – 1)2
⇒ f(0) = 0
⇒ f(1) = 1(1 – 1)2
⇒ f(1) = 02
⇒ f(1) = 0
∴ f(0) = f(1), Rolle’s theorem applicable for function ‘f’ on [0,1].
Let’s find the derivative of f(x):
⇒
Differentiating using UV rule,
⇒
⇒ f’(x) = ((x – 1)2×1) + (x×2×(x – 1))
⇒ f’(x) = (x – 1)2 + 2(x2 – x)
⇒ f’(x) = x2 – 2x + 1 + 2x2 – 2x
⇒ f’(x) = 3x2 – 4x + 1
We have f’(c) = 0 cϵ(0,1), from the definition given above.
⇒ f’(c) = 0
⇒ 3c2 – 4c + 1 = 0
⇒
⇒
⇒
⇒
⇒ ϵ(0,1)
∴ Rolle’s theorem is verified.