Find the local extrema of $y=x^3$
Find the local extrema of $y=x^3$.
We know that the $x=0$ is the turning point so $(0,0)$ is the extrema point for $y=x^3$, but how do we know that it is local minima or maxima? For quadratic equations with two roots it is easy to find which one is the minima or maxima. But, here it has only one root.
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$\begingroup$Your function does not have any local extreme points. It is a strictly monotonously growing function (if $x<y$, then $x^3<y^3$).
It is true that its derivative ($y'=3x^2$) had a zero at $x=0$, but this is not enough to guarantee there will be a local extreme point because $y''(0)$ is also zero (if it were positive, we'd have a minimum while it it were negative, we'd have a maximum).
$\endgroup$ $\begingroup$It is neither local minimum nor maximum since for any $x>0$ we have $x^3>0$ and for any $x<0$ we have $x^3<0$. Another way to conclude this is to notice that $x\mapsto x^3$ is strictly increasing function, so it can't have local extrema.
$\endgroup$ 2 $\begingroup$A differentiable function has a local extremum at $x_0$ exactly when its derivative changes its sign at $x_0$. In this case, the derivative corresponds to $3x^2$, which is non-negative for each $x$, hence never changes its sign.
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