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The minimum point of the function ($x^3$ / $3$)- $x$ is at
a) x= 1

b) x = -1

c) x = 0

d) x = 1/ √ 3
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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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