A chained inequality means an intersection · source reference

Complete question

Without a graphing calculator solve $x^2/(x-1)\le4x\le8$.

[2011/MJC/Prelim/I/6a]

Printed working

The domain excludes $x=1$. Split the chained inequality into two conditions joined by and:

$$\begin{aligned} \frac{x^2}{x-1}\le4x\le8 &\Rightarrow\frac{x^2}{x-1}-4x\le0\quad\text{and}\quad4x\le8\\ &\Rightarrow\frac{x^2-4x(x-1)}{x-1}\le0\quad\text{and}\quad x\le2\\ &\Rightarrow\frac{x^2-4x^2+4x}{x-1}\le0\quad\text{and}\quad x\le2\\ &\Rightarrow\frac{4x-3x^2}{x-1}\le0\quad\text{and}\quad x\le2\\ &\Rightarrow\frac{x(4-3x)}{x-1}\le0\quad\text{and}\quad x\le2\\ &\Rightarrow x(4-3x)(x-1)\le0,\;x\ne1\quad\text{and}\quad x\le2. \end{aligned}$$

The cubic sign sketch gives $0\le x<1$ or $x\ge\tfrac43$. Intersect this union with $x\le2$ on the printed number line. Taking the intersection, the solution is

$$0\le x<1\quad\text{or}\quad\frac43\le x\le2.$$

Teacher margin notes

Find values of $x$ that satisfy both inequalities, so take the intersection between the two ranges of values of $x$ that satisfy the inequalities; thus use “and”.

To find the intersection between the ranges of values of $x$, use a number line.

Source diagrams

Provenance

Question page references: 2. Solution page references: 9. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.