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The value of c, for which $f(x)=x(x-3)^2, 0 ≤x≤ 3 $ satisfies Rolle's theorem is : |
0 1 2 3 |
1 |
$f(x)=x(x-3)^2$ 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 |
