Rational inequalities with a sign table

Text

Solve using a sign table, studying graphically the sign of each factor:

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

Solution

(a) $x^2+x+1>0$ always ($\Delta<0$) and $(x-2)^4>0$ for $x\neq2$: the sign equals that of $\dfrac{x}{(x-3)^3}$, i.e. of $x(x-3)$. Hence $\le 0$ for $0\le x\le 3$, excluding $x=2,3$ (zeros of the denominator): $\;0\le x<3,\ x\neq 2$.

(b) Bring everything to one side. With $\dfrac{x+1}{2-x}=-\dfrac{x+1}{x-2}$ we get $\dfrac{6(x-1)}{x(x-2)}\le 0$. Sign of $\dfrac{x-1}{x(x-2)}$: solution $x<0 \ \vee\ 1\le x<2$.

(c) $\dfrac{(2-x)(2+x)}{x}>0$. Sign study with breakpoints $-2,0,2$: solution $x<-2 \ \vee\ 0<x<2$.

(d) $\dfrac{(1-x)^2}{(x-1)^4}=\dfrac{1}{(x-1)^2}>0$ for every $x\neq1$: never $\le 0$. No solution.

$\boxed{\text{(a) } 0\le x<3,\,x\neq2;\ \ \text{(b) } x<0\vee 1\le x<2;\ \ \text{(c) } x<-2\vee 0<x<2;\ \ \text{(d) } \varnothing}$