I want to find whether $x\mapsto x^2 + x$ is uniformly continuous on $(0,\infty)$, I know that $x\mapsto x^2$ is not uniformly continuous on this interval, however am having difficulty grasping whether this function is. This is what I have done so far I have made the function $$|x^2 +x -y^2 +x|<\varepsilon$$ where I made it $$|(x+y)(x-y)(1)|<\varepsilon.$$
Am I on the right track?
On
Let $f(x)=x^2+x$ for $x\in(0,\infty)$. Take $\varepsilon=1$. Let $\delta>0$ be given. Choose $M$ such that $$M > \frac1\delta +\delta+\frac12 .$$ Then $\left|M - (M+\frac\delta2)\right|=\frac\delta2<\delta$, but \begin{align}\left|f(M) - f(M+\frac\delta2)\right| &= \left(M+\frac\delta2\right)^2 - (M+\frac\delta2) - (M^2 - M)\\ &= M^2 + M\delta + \frac{\delta^2}4 - M-\frac\delta 2 - M^2 + M\\ &= M\delta +\frac{\delta^2}4 - \frac\delta2\\ &> \left(\frac1\delta+\delta+\frac12\right)\delta + \frac{\delta^2}4-\frac\delta2\\ &= 1+\frac{5\delta^2}4 > 1. \end{align} Hence, $f$ is not uniformly continuous.
On
If $f'$ is monotonic increasing and unbounded on $(0,\infty)$, then the function $f$ cannot be uniformly continuous. This should be intuitively true, just looking at the graph, so let's prove it. Without loss of generality, $f'>0$. Pick any $\epsilon>0$ and suppose there exists a $\delta>0$ such that
$$|x-y|<\delta\,\Rightarrow\,|f(x)-f(y)|<\epsilon.$$
Then in particular $|f(x+\frac{1}{2}\delta)-f(x)|<\epsilon$ for all $x$, so it's bounded. But
$$\left|f\left(x+\frac{1}{2}\delta\right)-f(x)\right|=\int_x^{x+\delta/2}f'(u)\,{\rm d}u>\frac{1}{2}\delta\, f'(x)\to\infty,$$
and this is a contradiction!
The sum of uniformly continuous functions is again an uniformly continuous function, and if you multiply the function by a scalar too. Note that the identity is uniformly continuous (it is Lipschitz). You know that $x^2$ is not uniformly continuous. If $x^2+x$ were uniformly continuous, we would have: $$x^2 = (x^2+x) - x$$ uniformly continuous, a contradiction. So $x^2+x$ is not uniformly continuous.