Is $y=x^{1/3}$ uniformly continuous on $[-1,1]$?

Question

When we check it by Heine-Cantor Theorem We conclude its uniform continuity however when we apply the 1st derivative test: that if the first derivative is bounded in the interval the function should be uniformly continuous; But this results in the function being not uniformly continuous. What am I missing?

Answer 1

The first derivative test says that if $f'$ is bounded, then $f$ is uniformly continuous. It does not say that the reverse implication holds.

Answer 2

As Another User says, you cannot reverse the implication of the first derivative test. Since it is relatively easy, it may be worth constructing an explicit relationship in the $\epsilon - \delta$ definition of uniform continuity to confirm it.

From the symmetry here, the tightest cases come when one of $x$ and $y$ is about $-\frac18 \epsilon^{3}$ and the other about $+\frac18 \epsilon^{3}$, suggesting we can use $\delta = \frac14\epsilon^{3}$. So we can say here that for any $\epsilon >0$ and any $x,y$ such that $|x-y|< \frac14\epsilon^{3}$ (not depending on $x$) we have $|x^{1/3}-y^{1/3}|< \epsilon$, satisfying the definition of uniform continuity.