Higher-degree and fractional inequalities

Text

Solve the following inequalities, using sign tables where needed:

• (a) $x^3-2x^2+x\le 0$
• (b) $x^4 \le 16$
• (c) $\dfrac{1}{x}+x\le \dfrac{5}{2}$
• (d) $\dfrac{(-2x+3)(4-x)}{x-1}<0$
• (e) $4-5x\le 8-3x$
• (f) $\dfrac{x^3+3x^2+2x}{x^2+x+1}\ge 0$

Solution

(a) $x^3-2x^2+x=x(x-1)^2\le 0$. Since $(x-1)^2\ge 0$, the sign depends on $x$: the inequality holds for $x\le 0$, plus the point $x=1$ where the product vanishes. Solution: $x\le 0 \ \vee\ x=1$.

(b) $x^4-16\le 0 \implies (x^2-4)(x^2+4)\le 0$. Since $x^2+4>0$, it reduces to $x^2-4\le 0$, i.e. $-2\le x\le 2$.

(c) $\dfrac{1}{x}+x-\dfrac52\le 0 \implies \dfrac{2x^2-5x+2}{2x}\le 0 \implies \dfrac{(2x-1)(x-2)}{2x}\le 0$. Breakpoints $0$ (excluded), $\tfrac12$, $2$. The sign table gives: $x<0 \ \vee\ \tfrac12\le x\le 2$.

(d) $\dfrac{(-2x+3)(4-x)}{x-1}<0$. Breakpoints $1$ (excluded), $\tfrac32$, $4$. Sign study: $x<1 \ \vee\ \tfrac32<x<4$.

(e) $4-5x\le 8-3x \implies -2x\le 4 \implies x\ge -2$.

(f) $\dfrac{x^3+3x^2+2x}{x^2+x+1}\ge 0$. The denominator $x^2+x+1$ is always positive ($\Delta<0$); the numerator is $x(x+1)(x+2)$. Studying the sign of the numerator gives: $-2\le x\le -1 \ \vee\ x\ge 0$.

$\boxed{\begin{aligned} &\text{(a) } x\le 0 \vee x=1; \quad \text{(b) } -2\le x\le 2; \quad \text{(c) } x<0 \vee \tfrac12\le x\le 2;\\ &\text{(d) } x<1 \vee \tfrac32<x<4; \quad \text{(e) } x\ge -2; \quad \text{(f) } -2\le x\le -1 \vee x\ge 0\end{aligned}}$