Using the triangle inequality to bound $\frac{x^3 + 3x + 1}{10-x^3}$ for $|x+1|<2$ How do I use the triangle inequality to bound the function $$f(x) = \frac{x^3 + 3x + 1}{10 - x^3}$$
on the interval $|x+1|<2$? I understand how the triangle inequality works, but using fractions with triangle inequality is confusing me.

Hint.
Use $\vert \vert x \vert -  \vert y \vert \vert \le \vert x-y \vert$ with $y=-1$. You'll get $\vert x \vert < 3$ with your hypothesis. Hence you can bound
$$\vert x^3 + 3x + 1 \vert \le \vert x \vert^3 + 3\vert x \vert + 1 < 37$$
Also for $x \le 0$ you have $-x^3 \ge 0$. Therefore for $x \le 0$: $0  \le \frac{1}{10 - x^3} \le \frac{1}{10}$. And for $x \ge 0$ as $|x+1|<2$ you have $0 \le x <1$ and also $0 \le x^3 <1$ hence $10-x^3 \ge 9$.
Finally you get:
$$\left\vert {(x^3 + 3x + 1)\over(10 - x^3)}\right\vert < \frac{37}{9}$$
