The value of c, for which f(x)=x(x-3)^2, 0 ≤x≤ 3 satisfies Rolle's th…

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

so $f(0)=f(3)=0$

so differentiating wrt x $f'(x)=(x-3)^2+2x(x-3)$

at some point $f'(x) = 0$

so $(x-3)(x-3+2x)$

$=3(x-3)(x-1)=0$

so at $x=1=C$ Rolle's theorem satisfied