Given $f:[0,8]\to \mathbb{R}$ be defined by $f(x)=x^{(1/3)}$ given that f is continuous prove that $f$ is uniformly continuous .Is $f'$ is bounded
my attempt: since $f'=\frac{1}{3}f^{(-2/3)}$ so $f'$ is not bounded but how to show that is uniformly continuous
It is not hard to prove that $f$ is continuous at $x=0$. Given $\epsilon>0$, there is a $\delta>0$ small enough such that $\delta<1$ and for all $x\in[0,8]$, if $x\leq\delta$, then $|f(x)|<\epsilon/2$.
And it is also not hard to prove that $f$ is continuous at $x=\delta$, so there is an $\eta>0$ small enough such that $\eta<\min\{\delta/4,(M+1)^{-1}\epsilon\}$ and for all $x\in[0,8]$, if $x\in[\delta-\eta,\delta+\eta]$, then $|f(x)-f(\delta)|<\epsilon/2$
Now $M:=\sup_{x\in[\delta/4,1]}|f'(x)|<\infty$. Then for all $x,y\in[0,8]$, if $|x-y|<\eta$, then we have four cases.
Case I) Both $x,y<\delta-\eta$. Then $|f(x)-f(y)|<2\cdot(1/2)\epsilon=\epsilon$
Case II) Both $x,y\in[\delta-\eta,\delta+\eta]$. Then $|f(x)-f(y)|\leq|f(x)-f(\delta)|+|f(y)-f(\delta)|<2\cdot(1/2)\epsilon=\epsilon$.
Case III) Only one, say, $x\in[\delta-\eta,\delta+\eta]$. Then $x\geq\delta-\eta>(3/4)\delta$, and also $y>x-\eta\geq\delta-2\eta>\delta/2$, so $x,y\in[\delta/4,1]$, and hence $|f(x)-f(y)|\leq M|x-y|\leq M(M+1)^{-1}\epsilon<\epsilon$.
Case IV) Both $x,y>\delta+\eta$. Then it is still $x,y\in[\delta/4,1]$, so $|f(x)-f(y)|\leq M|x-y|<\epsilon$.
In either case, we have $|f(x)-f(y)|<\epsilon$, the result follows.