My book states both of the functions 1 and 2 have no intervals. for 1) $f'(x)=(1/3)x^{-2/3}$ which is >0 for every real value of x and 2) $f'(x)=(2/3)x^{-1/3}$ which is <0 for (-∞,0) and >0 for (0,∞)
but both of the first derivatives are undefined at x=0.
So, can we say that $f(x)=\sqrt[3]{x}$ is increasing on (-∞, 0) U (0, +∞)
and $f(x)=2+x^{2/3}$ is decreasing on (-∞,0) and increasing on (0,∞)
thus we get intervals. Am i correct?
1) Correct. The fact that the derivative at $0$ is undefined doesn't preclude monotonicity. You know that $f(x)=\sqrt[3]{x}$ is strictly increasing on the open intervals $(-\infty, 0)$, $(0, \infty)$, and also know that $f(0) \lt f(x)$ for $\forall x \gt 0$ and $f(0) \gt f(x)$ for $\forall x \lt 0$. This makes $f(x)$ strictly increasing on the whole $\mathbb{R} = (-\infty,\infty)$.
2) Technically correct, but both intervals of monotonicity should include $\{0\}$, as $f(x)$ is strictly increasing on $[0,\infty)$ and strictly decreasing on $(-\infty,0]$. See previous step why.
[ EDIT ] The question was edited since I posted my answer. After the latest edit, my former
1) Correct.would rather now be1) Technically correct, but.... It is technically correct that $\sqrt[3]{x}$ is increasing on $(-\infty,0) \cup(0,\infty)$ but that's a weaker statement than $\sqrt[3]{x}$ being increasing on the whole of $\mathbb{R}$. Just like for example saying that $\sqrt[3]{x}$ is increasing on $(1,2) \cup (3,5)$ is technically correct, but is not the best or most relevant assessment of the monotonicity of $\sqrt[3]{x}$.