Fractional inequalities with sign analysis

Text

Solve:

• (a) $\dfrac{3-x}{1+x}\le 0$
• (b) $\dfrac{(5-x)(x^2+4)}{x^2-4}\ge 0$
• (c) $\dfrac{1}{2-x}-\dfrac{3x}{(2-x)^2}\ge \dfrac{x}{(x+1)(2-x)}$

Solution

(a) Domain: $x\neq -1$. Critical points: $x=3$ (numerator), $x=-1$ (denominator).

	$x<-1$	$-1<x<3$	$x>3$
$3-x$	$+$	$+$	$-$
$1+x$	$-$	$+$	$+$
ratio	$-$	$+$	$-$

The ratio is $\le 0$ for $x<-1\ \vee\ x\ge 3$ (at $x=3$ the numerator vanishes).

(b) Since $x^2+4>0$, the inequality is equivalent to $\dfrac{5-x}{(x-2)(x+2)}\ge 0$, with domain $x\neq \pm 2$. Critical points: $-2,\ 2,\ 5$.

	$x<-2$	$-2<x<2$	$2<x<5$	$x>5$
$5-x$	$+$	$+$	$+$	$-$
$(x-2)(x+2)$	$+$	$-$	$+$	$+$
ratio	$+$	$-$	$+$	$-$

Solution: $x<-2\ \vee\ 2<x\le 5$ (at $x=5$ the numerator vanishes).

(c) Domain: $x\neq 2,\ x\neq -1$. Bringing everything over the common denominator $(2-x)^2(x+1)$: $\frac{(2-x)(x+1)-3x(x+1)-x(2-x)}{(2-x)^2(x+1)}\ge 0.$ The numerator is $(-x^2+x+2)+(-3x^2-3x)+(x^2-2x)=-3x^2-4x+2$, hence $\frac{-3x^2-4x+2}{(2-x)^2(x+1)}\ge 0 \iff \frac{3x^2+4x-2}{(2-x)^2(x+1)}\le 0.$ Roots of the numerator: $x=\dfrac{-2\pm\sqrt{10}}{3}$ (numerically $\approx -1.721$ and $\approx 0.387$). Since $(2-x)^2>0$ (with $x\neq 2$), the sign depends on $\dfrac{3x^2+4x-2}{x+1}$. Ordered critical points: $\dfrac{-2-\sqrt{10}}{3},\ -1,\ \dfrac{-2+\sqrt{10}}{3}$.

	$x<\frac{-2-\sqrt{10}}{3}$	$\frac{-2-\sqrt{10}}{3}<x<-1$	$-1<x<\frac{-2+\sqrt{10}}{3}$	$x>\frac{-2+\sqrt{10}}{3}$
$3x^2+4x-2$	$+$	$-$	$-$	$+$
$x+1$	$-$	$-$	$+$	$+$
ratio	$-$	$+$	$-$	$+$

The ratio is $\le 0$ for $x\le \dfrac{-2-\sqrt{10}}{3}\ \vee\ \Big(-1<x\le \dfrac{-2+\sqrt{10}}{3}\Big)$, with $x\neq 2$ (which lies outside these intervals anyway).

$\boxed{\begin{aligned} &\text{(a) } x<-1 \vee x\ge 3; \quad \text{(b) } x<-2 \vee 2<x\le 5;\\ &\text{(c) } x\le \tfrac{-2-\sqrt{10}}{3}\ \vee\ \Big(-1<x\le \tfrac{-2+\sqrt{10}}{3}\Big)\end{aligned}}$