Question

f(x)=x(x-3)² x ∈ [0,3],Verify Rolle's theorem

Answer 1

Answer:

The suitable value of c is 1

Step-by-step explanation:

F(x)=x(x-3)² x ∈ [0,3],Verify Rolle's theorem

Given:

$f(x)=x(x-3)^2$

$f(x)=x(x^2+9-6x)$

$f(x)=x^3+9x-6x^2$

$f'(x)=3x^2+9-12x$

$f'(x)=3(x^2-4x+3)$

since f(x) is a polynomial,

f(x) is continuous on [0,3]

f(x) is differentiable in (0,3)

Also, f(0)=0 and f(3)=0

$\implies\:f(0)=f(3)$

$\therefore\text{Conditions of rolle's theorem are satisfied}$

Now,

$f'(c)=0$

$\implies\:c^2-4c+3=0$

$\implies\:(c-1)(c-3)=0$

$\implies\:c=1,3$

But 3∉(0,3) and 1∈(0,3)

$\therefore\text{The suitable value of c is 1}$

$\text{Hence, Rolle's theorem verified}$

Answer 2

Answer:

Step-by-step explanation: