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$Let f(x)=x^{3}/3 - x$;

${f(x)}'= 2x^{2}-1$;

Now Putting ${f(x)}'=0$;

Solving we get $x=1$,$-1$;

${f(x)}''$ is greater than zero for $x=1$

which implies $x=1$ is a point of minima.

And the minimum value of $f(x)=-2/3$

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