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 – 4)2 on [0, 4]
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 – 4)2
⇒ f(0) = 0
⇒ f(4) = 4(4 – 4)2
⇒ f(4) = 4(0)2
⇒ f(4) = 0
∴ f(0) = f(4), Rolle’s theorem applicable for function ‘f’ on [0, 4].
Let’s find the derivative of f(x):
\(\Rightarrow f'(x) = \frac{d(x(x-4)^2)}{dx}\)
Differentiating using UV rule,
We have f’(c) = 0 c∈(0,4), from the definition given above.
∴ Rolle’s theorem is verified.