Question

(iv) x3 - 6x2 +11% -6 by x2 +2​

Answer 1

Answer:

Step-by-step explanation:

heck the applicability of Rolle’s Theorem for the function:

f(x) = 1

−2

on [0,2].

Solution

Given f(x)= 1

−2

on [0,2].

Clearly, f(x) is not defined at x=2∈ [0,2]

∴ () is not continuous at x=2∈ [0,2]

∴ () is not continuous on [0,2].

Since, the conditions of c are not satisfied, so Rolle’s Theorem is not applicable for

f(x)= 1

−2

on [0,2].

Question 2

Verify Rolle’s Theorem for the function f(x)=x(x-3)2

in the interval [0,3].

Solution

We have f(x)=x(x-3)2 =x3

-6x2+9x

(i)f is continuous in the closed interval [0,3],

(ii)f is differentiable in the open interval (0,3) and

(iii)f(0)=0=f(3)

Thus all the conditions of the Rolle’s Theorem are satisfied.

Then there exists at least one real number c in (0,3)