Inventing an inequality given its solution set

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For each of the following solution sets, invent an inequality having it as its solution:

• (a) $2\le x \le 3$
• (b) $x\neq 4$
• (c) $x\le -4 \ \vee\ 0<x\le 3$

Solution

Work backwards: pick the breakpoints (zeros and excluded values) and build a product/quotient with the desired sign.

(a) A $\le 0$ with zeros at $2$ and $3$ and upward concavity: $(x-2)(x-3)\le 0$.

(b) Exclude only $x=4$: $(x-4)^2>0$.

(c) Breakpoints $-4,0,3$ with $0$ excluded (denominator). Build $\dfrac{(x+4)(x-3)}{x}\le 0$: the sign table gives exactly $x\le-4 \ \vee\ 0<x\le 3$ (zeros $-4,3$ included, $0$ excluded).

$\boxed{\text{(a) } (x-2)(x-3)\le 0;\quad \text{(b) } (x-4)^2>0;\quad \text{(c) } \tfrac{(x+4)(x-3)}{x}\le 0}$