Verify Rolle’s theorem for each of the following functions:

Show that satisfies Rolle’s theorem on [0, 5] and that the value of c is (5/3)

Condition (1):

Since, f(x)=x(x-5)² is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x(x-5)² is continuous on [0,5].

Condition (2):

Here, f’(x)= (x-5)²+ 2x(x-5) which exist in [0,5].

So, f(x)= x(x-5)² is differentiable on (0,5).

Condition (3):

Here, f(0)= 0(0-5)²=0

And f(5)= 5(5-5)²=0

i.e. f(0)=f(5)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,5) such that f’(c)=0

i.e. (c-5)²+ 2c(c-5)=0

i.e.(c-5)(3c-5)=0

i.e. or c=5

Value of

Thus, Rolle’s theorem is satisfied.