FP2.1 - Inequalities

Learning objectives

—FP2.1.1 - Manipulation and solutionThe manipulation and solution of The solution of inequalities such as algebraic inequalities and 1 x inequations, including those >, | x2 − 1 | > 2(x + 1). x − a x − b involving the modulus sign.

An algebraic inequality is solved by finding every point where its truth value can change, then determining which intervals satisfy the original relation. Denominator zeros and modulus boundaries must be recorded before simplifying.

Form	Safe algebraic route
rational inequality	Move all terms to one side, combine them into $N(x)/D(x)$ , and mark zeros of both $N$ and $D$ . Test the sign on each resulting interval.
$\|F(x)\|<G(x)$	Require $G(x)>0$ , then solve $-G(x)<F(x)<G(x)$ . For $\le$ , require $G(x)\ge0$ .
$\|F(x)\|>G(x)$	Where $G(x)<0$ the inequality is automatic; where $G(x)\ge0$ , solve $F(x)>G(x)$ or $F(x)<-G(x)$ . Use the analogous inclusive conditions for $\ge$ .
piecewise modulus	Alternatively split at every zero of the expression inside $\|\ \|$ , replace it by the correct signed expression on each interval, and intersect each result with that interval.

For example, $\frac1{x-1}>\frac{x}{x+2}\quad\Longleftrightarrow\quad\frac{-x^2+2x+2}{(x-1)(x+2)}>0.$ The numerator is zero at $x=1\pm\sqrt3$ ; the denominator is zero at $x=-2,1$ . A sign chart across these four ordered critical points gives $-2<x<1-\sqrt3\quad\text{or}\quad1<x<1+\sqrt3.$ The denominator points are excluded even though they separate sign intervals.

For the official model inequality $|x^2-1|>2(x+1)$ , the right side is negative when $x<-1$ , so that whole interval works automatically. For $x\ge-1$ , solve $x^2-1>2(x+1)\quad\text{or}\quad x^2-1<-2(x+1).$ These become $(x-3)(x+1)>0$ or $(x+1)^2<0$ , giving $x>3$ . Hence the full solution is $x<-1\quad\text{or}\quad x>3.$

Never multiply an inequality by an expression whose sign is unknown: the inequality direction might need to reverse. Keep exact critical values, test every interval, exclude every point where the original expression is undefined, and include a numerator or equality boundary only when the original inequality permits equality.

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