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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?

Sahil
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1 Answers1

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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 be 1) 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 $\sqrt[3]{x}$.
dxiv
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  • For 2),you have included x=0 on both increasing and decreasing interval. So particularly at x=0 we get a spike in the curve and how can we determine whether the function is increasing or decreasing at x=0 since first derivative is undefined at x=0. – Sahil Nov 22 '16 at 05:06
  • how can a function both increase and decrease at a same point? in reference to x=0 for second function. – Sahil Nov 22 '16 at 05:08
  • @ManishYadav A function increases or decreases on an interval not at one point. You may want to read my answer to a similar question at Monotonicity of quadratic function at its vertex?. – dxiv Nov 22 '16 at 05:10
  • i mean to say that the particular point is contained in an arbitrary interval – Sahil Nov 22 '16 at 05:12
  • @ManishYadav Any particular point can be contained in many intervals. This doesn't change the fact that, as I said, a function increases or decreases on an interval not at one point. Do you disagree that $2+\sqrt[3]{x^2}$ is strictly increasing on $[0,\infty)$ and *also* strictly decreasing on $(-\infty,0]$? – dxiv Nov 22 '16 at 05:16
  • I agree to all the things you said, except the existence of x=0 on both the intervals. Your statement says at x=0, the second function is both increasing and decreasing which is impossible. According to me, it should have been strictly decreasing on (−∞,0) and strictly increasing on (0,+∞). – Sahil Nov 22 '16 at 05:21
  • @ManishYadav Again, you insist on considering monotonicity at a point. Consider that the very definition of monotonicity requires $2$ distinct points. Back to the issue at hand, if you disagree that $2 + \sqrt[3]{x^2}$ is strictly increasing on the half-closed interval $[0,\infty)$ then please make your case why that's wrong, starting from the definition of monotonicity. – dxiv Nov 22 '16 at 05:26
  • Let us continue this discussion in chat. – Sahil Nov 22 '16 at 05:27
  • Sorry, I don't do chats. Nothing personal or question related, just that I don't generally chat at all. – dxiv Nov 22 '16 at 06:06