Proof $f(x) = x^2$ is uniformly continuous in $[0,17]$

Question

Let $\epsilon > 0$ i need to find $\delta > 0$ such that $\forall x,y \in [0,17]$ if $|x-y| < \delta$ $\Rightarrow$ $|f(x) - f(y)| < \epsilon$.

$|f(x) - f(y)| = |x^2 - y^2| = |(x+y)(x-y)| = |(x+y)| |(x-y)|$ ( because $x,y > 0$)

$0 \le x+y \le 34$ (due to the domain), Therefore $\Rightarrow$ $|(x-y)| \le 34 \cdot (x-y)$

Hence, we demand - $$\delta = \frac{\epsilon}{34}$$

And now, $\forall x,y \in [0,17]$ such that $|(x-y)| < \delta$ $\Rightarrow$ $|f(x) - f(y)| < \epsilon$ so $f(x)$ is uniformly continuous in $[0,17]$.

Did i do it right ? Thanks.

Answer 1

Almost. You justify the equaliy $\bigl|(x+y)(x-y)\bigr|=\bigl|(x+y)\bigr|\bigl|(x-y)\bigr|$ with the argument “because $x,y>0$”. Actually, this is always true. Where you need to use that $x,y\geqslant 0$ is in $|x+y|=x+y$. And you are wrong when you write that $\bigl|(x-y)\bigr|\leqslant 34(x-y)$. What you proved was that $|x^2-y^2|\leqslant34|x-y|$.