Uniform continuity of $f(x)=x^{2/3}\log x$ on $[0, \infty)$

Question

Let $f$ be a real valued function on $[0,\infty)$ defined by

$$f(x)= \begin{cases} x^{{2}/{3}}\log x & \text{ if } x>0 \\ 0 & \text{ if } x= 0 \end{cases}$$

Then which of the following is true?

A. $f$ is discontinuous at $x=0$.

B. $f$ is continuous on $[0,\infty)$, but not uniformly continuous on $[0,\infty)$.

C. $f$ is uniformly continuous on $[0,\infty)$.

D. $f$ is not uniformly continuous on $[0,\infty)$, but uniformly continuous on $[1,\infty)$.

I could see that $f(x)=x^{\frac{2}{3}}\log x$ is continous at $x=0$

But for uniform continuity I thought boundedness of derivative would help

But I found $f'(x)=\frac{2}{3}x^{-\frac{1}{3}}\log x + x^{-\frac{1}{3}}$

This $f'(x)$ is not bounded so i can not say any thing about uniform contunuity.

Please help me to clear this.

Thank you :)

P.S : This was already asked before (Uniform continuity of $f(x)=x^{\frac{2}{3}}\log x$ some other person but it was not having enough details so it got closed. I tried to edit that question but someone said I should not edit OP to add more details. I was afraid that may be against the rules of MSE. So, I thought i should ask this separately with my own ideas.

Answer 1

You proved that the function $f$ is continuous on $[0,1]$, hence uniformly continuous on $[0,1]$. The expression of $f'$ you provided, plus the inequality $\log u\leqslant u$ applied to $u=x^{1/3}$, yield $0\leqslant f'(x)\leqslant3$ for every $x\geqslant1$, hence $f$ is uniformly continuous on $[1,+\infty)$. Ergo.