Respuesta :
Answer:
Rolle's theorem can be applies over the interval [0,2] for the given function f(x).
Step-by-step explanation:
We are given the following in the question;
[tex]f(x) = x(x-2)^2 \text{ in }[0,2][/tex]
The Rolle's theorem states that is
- f(x) is continuous on the interval [a,b]
- f(x) is differentiable on the interval (a,b)
- f(a) = f(b)
then, there exist c in the interval (a,b) such that
[tex]f'(c) = 0[/tex]
Condition:
Continuity of f(x)
Since f(x) is a polynomial function, it is continuous i the given interval [0,2].
Differentiability of f(x)
Since f(x) is a polynomial function, it is continuous i the given interval (0,2).
Equality of f(x)
[tex]f(0) = 0(0-2)^2 = 0\\f(2) = 2(2-2)^2 = 0\\f(0) = f(2)[/tex]
Thus, Rolle's theorem can be applies over the interval [0,2] for the given function f(x).
