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Q.
The local maximum value of $x(1-x)^2, 0 \leq x \leq 2$ is-
Application of Derivatives
Solution:
$f(x)=x(1-x)^2$
$f^{\prime}(x)=-2 x(1-x)+(1-x)^2=0$
$(1-x)(-2 x+1-x)=0$
$\Rightarrow (1-x)(1-3 x)=0, x=1,1 / 3$
Local max. at $x=1 / 3$
which is $f(x)=\frac{1}{3}\left(1-\frac{1}{3}\right)^2=\frac{4}{27}$
