# Nonlinear inequalities in one-dimension

**Inequalities** involving numerical functions can be rewritten as $$$f(x)\lt 0,\qquad f(x)\le 0$$$

The inequality $$g(x)\gt 1$$ can be written as $$1-g(x)\lt 0$$.

For a **quadratic function** with $$a\gt 0$$ and discriminant $$\Delta = b^2-4ac\gt0$$, the roots are $$\displaystyle x_\pm = \frac{-b\pm\sqrt{\Delta}}{2a}$$. So, we have $$$ f(x) = ax^2 + bx + c = a(x-x_-)(x-x_+).$$$ So $$f\lt 0$$ on $$(x_-,x_+)$$ and $$f \ge 0$$ on $$(-\infty,x_-]\cup[x_+,+\infty)$$.

To compute the sign of a function, we decompose it as the **product or ratio of simple functions**. Take $$f=g/h$$, with $$g$$ and $$h$$ having distinct non stationary zeros (if $$g(x)=0$$ then $$g'(x)\ne0$$). Then, $$f$$ **alternates sign at the roots of $$g$$ and $$h$$**. Its limit at $$\pm\infty$$ gives the sign at $$\pm\infty$$.

$$f(x)=(x^2-1)/x$$ changes sign at $$-1$$, $$0$$ and $$1$$. Its limit at $$-\infty$$ is $$-1$$. Hence it is strictly negative on $$(-\infty,-1)\cup(0,1)$$.