$f(x)=x^\alpha$ uniformly continuous on [$0,\infty)$? using MVT in spivak's calculus, there is a question asking 

for which of the following values of $\alpha$ is the function $f(x)=x^\alpha$ uniformly continuous on [$0,\infty)$? : $\alpha = 1/3, 1/2, 2, 3$

I know that $x^\alpha$ is uniformly continuous on [$0,\infty)$ for $\alpha = 1/3, 1/2$ but not for $\alpha = 2,3$ but this is not my interest.
it was brought up in lecture that a proof of the result for general $\alpha \in [0,\infty)$ will be much easier after we learned the mean value theorem and now that i know what MVT is, how does one prove this with using MVT?
All I know is that $x^\alpha$ is uniformly continuous on [$0,\infty)$ for $0 \le \alpha \le 1$ and not for $\alpha > 1$, and the definition of MVT is, $f$ is continuous on [$a,b$] and diff on ($a,b$) then there exists $x \in (a,b)$ such that $f'(x) = \frac{f(b)-f(a)}{b-a}$
 A: 
CASE $1$: $\alpha >1$

Let $f(x)=x^\alpha$ for $x\in [0,\infty)$ and $\alpha>1$.  We wish to determine whether $f$ is uniformly continuous on $[0,\infty)$.  
From the mean value theorem, there exists a number  $x^*\in(x_1,x_2)$ such that 
$$|f(x_2)-f(x_1)|=|f'(x^*)|\,|x_2-x_1|=\alpha(x^*)^{\alpha-1}|x_2-x_1|$$
For $\epsilon=\alpha/2$, choose any $\delta>0$, however small.  Then, for $x_1=1/\delta^{1/(\alpha-1)}$ and $x_2=\frac12 \delta +1/\delta^{1/(\alpha-1)}$, we see that
$$|f(x_2)-f(x_1)|=|f'(x^*)|\,|x_2-x_1|=\alpha(x^*)^{\alpha-1}\delta/2\ge \alpha/2=\epsilon$$

Therefore, $f(x)=x^\alpha$ fails to be uniformly continuous on $[0,\infty)$ when $\alpha>1$.



CASE $2$: $\alpha < 1$ and $x_1>x_0>0$

Then given any $\epsilon>0$
$$|f(x_2)-f(x_1)|=\alpha(x^*)^{\alpha -1}|x_2-x_1|\le \alpha(x_0)^{\alpha-1}|x_2-x_1|<\epsilon$$
whenever $|x_2-x_1|<\delta =\frac{x_0^{1-\alpha}}{\alpha}\,\epsilon$.

Therefore, $f$ is uniformly continuous for $x\in [x_0,\infty)$ for all $x_0>0$ and $\alpha < 1$.



CASE $3$: $\alpha < 1$ and $x_1=0$

Here, we have simply 
$$|f(x_2)-f(x_1)|=|f(x_2)|=x_2^\alpha<\epsilon$$
whenever $|x_2-x_1|=x_2<\delta =\epsilon^{1/\alpha}$.

Therefore, $f$ is continuous at $x=0$.  Putting this together with the result from Case $2$, we see that $f$ is uniformly continuous on $[0,\infty)$ for $\alpha < 1$.



CASE $4$: $\alpha = 1$

Obviously, given $\epsilon>0$, $|f(x_2)-f(x_1)|=|x_2-x_1|<\epsilon$, whenever $|x_2-x_1|<\delta=\epsilon$.
A: MVT doesn't seem appropriate here directly, but you might show that the functions $x^2$ and $x^3$ become arbitrarily steep as $x \rightarrow \infty$ - this would show that they are not uniformly continuous. 
A: For a function to be uniformly continuous, we need to limit its slope for any given interval along the domain: $[x-\delta,x+\delta]$]. If the derivative of a function, $\frac{df(x)}{dx}$, approaches infinity without ever returning, no fixed $\delta$ can be determined that will remain uniform on $(-\infty,\infty)$, such as with $f(x)=x^n,c^x,\frac1x$.
However, if the slope of a function, $\frac{df(x)}{dx}$, goes to infinity "reversibly", or stays within certain parameters, it may be uniformly continuous. For $f(x)=x^\alpha$ when $\alpha=\frac13,\frac12$, the function can be re-written as $f(y)=y^\frac1\alpha$, which is uniformly continuous on $[x^*-\delta,x^*+\delta]$. As $f(x)$ approaches infinity, the derivatives diminish in amplitude, and thus, $f(x)$ is uniformly continuous.
