Question

The minimum point of the function $f\left(x\right)=\left(x^3/3\right)-x$ is at

• A. $x=1$
• B. $x=-1$
• C. $x=0$
• D. $x=\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $x=1$

$f\left(x\right)=\frac{x^3}{3}-x$
$f^{\prime }x=x^2-1$
for minima or minima, $f^{\prime }\left(x\right)=0$
$x^2-1=0$
$x=\pm 1$
for minima, $f^{\prime \prime }\left(x\right)>0$
$f^{\prime \prime }\left(x\right)=2x$
$f^{\prime \prime }\left(x\right)$ (at $x=1$) is $2>0$
Thus, minima at $x=1$
$f^{\prime \prime }\left(x\right)$ (at $x=-1$) is $-2<0$
Thus, maxima at $x=-1$