Integer inequalities of various degrees

Text

Solve:

• (a) $3x-2\le x+3$
• (b) $-x^2+x+4\le 0$
• (c) $-x^3+x^2+4x\ge 0$
• (d) $(x+1)^2(x-2)(3-2x)\le 0$

Solution

(a) $3x-2\le x+3 \implies 2x\le 5 \implies x\le \tfrac52$.

(b) Changing sign: $-x^2+x+4\le 0 \iff x^2-x-4\ge 0$. The roots are $x=\dfrac{1\pm\sqrt{1+16}}{2}=\dfrac{1\pm\sqrt{17}}{2}$. The parabola opens upward, so it is $\ge 0$ outside the roots: $x\le \dfrac{1-\sqrt{17}}{2}\ \vee\ x\ge \dfrac{1+\sqrt{17}}{2}.$

(c) Changing sign: $-x^3+x^2+4x\ge 0 \iff x^3-x^2-4x\le 0 \iff x(x^2-x-4)\le 0$. The critical points are $x=0$ and the roots $x=\dfrac{1\pm\sqrt{17}}{2}$ (numerically $\approx -1.56$ and $\approx 2.56$).

	$x<\frac{1-\sqrt{17}}{2}$	$\frac{1-\sqrt{17}}{2}<x<0$	$0<x<\frac{1+\sqrt{17}}{2}$	$x>\frac{1+\sqrt{17}}{2}$
$x$	$-$	$-$	$+$	$+$
$x^2-x-4$	$+$	$-$	$-$	$+$
product	$-$	$+$	$-$	$+$

The product is $\le 0$ for $x\le \dfrac{1-\sqrt{17}}{2}\ \vee\ 0\le x\le \dfrac{1+\sqrt{17}}{2}$.

(d) The factor $(x+1)^2\ge 0$ always, and vanishes at $x=-1$ (where the product is zero, so $\le 0$ holds). For $x\neq -1$ the sign depends on $(x-2)(3-2x)\le 0$. The roots are $x=2$ and $x=\tfrac32$; this product opens downward (leading coefficient $-2$), so it is positive between the roots and $\le 0$ outside: $x\le \tfrac32\ \vee\ x\ge 2$. The point $x=-1$ is already included.

$\boxed{\begin{aligned} &\text{(a) } x\le \tfrac52; \quad \text{(b) } x\le \tfrac{1-\sqrt{17}}{2} \vee x\ge \tfrac{1+\sqrt{17}}{2};\\ &\text{(c) } x\le \tfrac{1-\sqrt{17}}{2} \vee 0\le x\le \tfrac{1+\sqrt{17}}{2}; \quad \text{(d) } x\le \tfrac32 \vee x\ge 2\end{aligned}}$