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# 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 $(-\infty,-1)\cup(0,1)$.

Sign of the function $f(x) = \frac{x^2-1}{x}$