Prove a certain cubic function is uniformly continuous on the unit interval

Question

Let $f:[0, 1] \to \mathbb{R}$, $f(x) = -x^3+2x$. Prove using the definition of uniform continuity that the function is uniformly continuous.

By definition $\forall \varepsilon >0$ $\exists \delta>0$ such that, $|f(x)-f(y)|< \varepsilon$ whenever $x,y, \in \mathbb{R}$ and $|x-y|< \delta$.

Using this I get $|-x^3+2x+y^3-2y| = |-x^3+y^3 +2x-2y|$. How should I continue from here? I'm a bit stuck.

Answer 1

The triangle inequality yields $$ |y^3-x^32+2x-2y|\leq |y^3-x^3|+2|x-y|<|y-x||y^2+yx+x^2|+2\delta<3\delta+2\delta=5\delta. $$ So given any $\epsilon>0$, let $\delta=\epsilon/5$

Answer 2

Hint

$$|-x^3+y^3+2x-2y|=|x-y||2+y^2+x^2+yx|\leq 5|x-y|.$$ Take $\delta =\frac{\varepsilon }{5},$ and thus $$|x-y|<\delta \implies |f(x)-f(y)|<\varepsilon .$$

Answer 3

The depart is $$ |-x^3+ 2x + y^3 -2y|= |-(x-y)(x^2 + xy + y^2) + 2(x-y)|= |x-y||2-x^2-xy-y^2| $$ now let's take a value $\epsilon$. If $$ |x-y||2-x^2-xy-y^2|<\epsilon $$ holds, then $$ |x-y|<\frac{\epsilon}{|2-x^2-xy-y^2|} $$ holds as well.

The minimal $\delta$ (such that the above inequality holds for all ) is found when the denominator for $x,y\in [0,1]$ is minimum, hence when $x,y=0$. Then we find $$ \delta=\epsilon/2.$$

Comparing with the other answers this seams to be the higher bound value for $\delta$ found so far.