Constructing an inequality given the sign table

Statement

Devise a (rational) inequality that produces the following sign table, where the last row represents the overall sign:

	$x<-3$	$x=-3$	$-3<x<2$	$x=2$	$2<x<4$	$x=4$	$x>4$
$f_1$	$+$	$0$	$-$	$-$	$-$	$-$	$-$
$f_2$	$-$	$-$	$-$	$\nexists$	$+$	$+$	$+$
$f_3$	$-$	$-$	$-$	$-$	$-$	$0$	$+$
total	$+$	$0$	$+$	$\nexists$	$-$	$0$	$+$

Hint: $f_1$ is a decreasing straight line (going from $+$ to $-$) with a zero at $-3$; $f_2$ is the denominator (marked by the double bar at $x=2$); $f_3$ is an increasing straight line with a zero at $4$.

Solution

We reconstruct the three factors from their behaviour:

• $f_1$ decreasing with a zero at $-3$: $f_1 = -x-3$ (positive for $x<-3$, negative afterwards).
• $f_2$ denominator with a key point at $2$ (point of non-existence): $f_2 = x-2$.
• $f_3$ increasing with a zero at $4$: $f_3 = x-4$.

One possible answer is $\boxed{\dfrac{(-x-3)(x-4)}{x-2} \le 0}$ that is, equivalently, $\dfrac{(x+3)(x-4)}{x-2} \ge 0$.

Links

Topics: Inequalities
Concepts: Key point · Rational inequality · Sign analysis · Sign table
Skills: Reasoning by cases · Solving inequalities
Exercise types: Solving an inequality