The wavy curve method

The rule "pos­i­tive out­side the roots, neg­a­tive between them" works for qua­drat­ics. For prod­ucts of three or more fac­tors, and for frac­tions, we need some­thing more gen­eral. The wavy curve method (also called the sign scheme or method of inter­vals) han­dles them all with one pic­ture: mark the crit­i­cal points on the num­ber line, and let the sign wave up and down across them. This les­son explains the method and the two spe­cial cases that need care: repeated fac­tors and points where a frac­tion is unde­fined.

Why signs change only at crit­i­cal points#

An expres­sion made of fac­tors like $(x - 1)$, $(x - 2)$, $(x - 3)$ mul­ti­plied or divided can only change sign where one of the fac­tors is zero. Between two neigh­bour­ing such points, every fac­tor keeps its sign, so the whole expres­sion keeps its sign. These points are called crit­i­cal points. So we need only find the sign in each stretch between crit­i­cal points.

The method#

1. Bring every­thing to one side, so that the inequal­ity com­pares an expres­sion with 0. Do not mul­ti­ply both sides by an expres­sion con­tain­ing $x$.
2. Fac­torise the top and bot­tom into lin­ear fac­tors, and arrange each fac­tor so that the coef­fi­cient of $x$ is pos­i­tive. (If you mul­ti­ply by $-1$ to do this, reverse the inequal­ity.)
3. Mark all the crit­i­cal points, the zeros of the top and of the bot­tom, on the num­ber line in increas­ing order.
4. Find the sign after the last (largest) crit­i­cal point. With every coef­fi­cient of $x$ pos­i­tive, it is $+$.
5. Move to the left. At each crit­i­cal point the sign changes if the fac­tor occurs an odd num­ber of times, and stays the same if it occurs an even num­ber of times. Drawn as a curve above the line for $+$ and below for $-$, the signs form a wave.
6. Read off the answer. Points where the expres­sion is zero (zeros of the top) are included for $\ge$ or $\le$. Points where the expres­sion is unde­fined (zeros of the bot­tom) are never included.

Exam­ple 1#

Solve $(x - 1)(x - 2)(x - 3) > 0$.

The crit­i­cal points are 1, 2, 3. To the right of 3 every fac­tor is pos­i­tive, so the sign is $+$. Mov­ing left, the sign changes at each point (each fac­tor occurs once):

We want $> 0$, so $x \in (1, 2) \cup (3, \infty)$.

Check with $x = 1.5$: $(0.5)(-0.5)(-1.5) = 0.375 > 0$. Cor­rect.

Exam­ple 2#

Solve $(x + 2)(x - 1)(x - 4) \le 0$.

The crit­i­cal points are $-2$, 1 and 4. The signs from right to left are $+, -, +, -$:

We want $\le 0$, so we take the neg­a­tive stretches and include the zeros: $x \in (-\infty, -2] \cup [1, 4]$.

Exam­ple 3#

Solve $(2 - x)(x + 3) > 0$.

The fac­tor $2 - x$ has a neg­a­tive coef­fi­cient of $x$. Write it as $-(x - 2)$ and mul­ti­ply the inequal­ity by $-1$, revers­ing it: $(x - 2)(x + 3) < 0$. The crit­i­cal points are $-3$ and 2, with signs $+, -, +$ from left to right. So $x \in (-3, 2)$.

Repeated fac­tors#

A fac­tor that appears an even num­ber of times, such as $(x - 1)^2$, is never neg­a­tive, so it does not change the sign of the expres­sion as $x$ passes its root.

Exam­ple 4#

Solve $(x - 1)^2(x - 3) \ge 0$.

Crit­i­cal points: 1 (from a squared fac­tor, so even) and 3 (odd). To the right of 3: $+$. Cross­ing 3, the sign changes to $-$. Cross­ing 1, it does not change, so it stays $-$.

The expres­sion is $\ge 0$ for $x \ge 3$, and also at $x = 1$, where it equals 0. So $x \in \{1\} \cup [3, \infty)$. For­get­ting the iso­lated point $x = 1$ is a com­mon error.

Frac­tions#

For a frac­tion, the zeros of the denom­i­na­tor are crit­i­cal points too, because the sign can change there. But the frac­tion is unde­fined at those points, so they are always excluded.

Exam­ple 5#

Solve $\displaystyle \frac{x - 1}{x + 2} < 0$, and then $\displaystyle \frac{x - 1}{x + 2} \ge 0$.

The crit­i­cal points are $-2$ (bot­tom) and 1 (top). The signs from left to right are $+, -, +$.

For $< 0$: $x \in (-2, 1)$.

For $\ge 0$: the pos­i­tive stretches together with the zero at $x = 1$, but not $x = -2$, where the frac­tion is unde­fined: $x \in (-\infty, -2) \cup [1, \infty)$.

Exam­ple 6#

Solve $\displaystyle \frac{(x - 1)(x - 2)^2}{x - 3} \le 0$.

At $x = 1$ and $x = 2$ the expres­sion is zero, and at $x = 3$ it is unde­fined. The fac­tor $(x - 2)^2$ is even.

To the right of 3: $+$. Cross­ing 3 (odd): $-$. Cross­ing 2 (even): stays $-$. Cross­ing 1 (odd): $+$.

We want $\le 0$: the two neg­a­tive stretches, the zeros at 1 and 2, but not 3. So $x \in [1, 3)$.

Exam­ple 7#

Solve $\displaystyle \frac{2x - 1}{x - 3} \ge 1$.

It is tempt­ing to mul­ti­ply both sides by $x - 3$, but its sign is unknown, so we can­not tell whether to reverse the inequal­ity. Instead bring every­thing to one side:

$$\displaystyle \frac{2x - 1}{x - 3} - 1 \ge 0 \implies \frac{2x - 1 - (x - 3)}{x - 3} \ge 0 \implies \frac{x + 2}{x - 3} \ge 0$$

Crit­i­cal points $-2$ and 3, signs $+, -, +$. So $x \in (-\infty, -2] \cup (3, \infty)$.

Fac­tors that never van­ish#

A fac­tor such as $x^2 + 1$ or $x^2 + x + 1$ is pos­i­tive for every $x$, so it can sim­ply be ignored (divided out) with­out affect­ing the sign.

Exam­ple 8#

Solve $(x^2 + 1)(x - 5) < 0$ and $x^3 - 4x > 0$.

Since $x^2 + 1 > 0$ always, the first is the same as $x - 5 < 0$: $x \in (-\infty, 5)$.

For the sec­ond, $x^3 - 4x = x(x - 2)(x + 2)$, with crit­i­cal points $-2$, 0, 2 and signs $-, +, -, +$ from left to right. So $x \in (-2, 0) \cup (2, \infty)$.

Prac­tice#

1. Solve $(x + 1)(x - 2)(x - 5) < 0$.
2. Solve $x(x - 3)(x + 4) \ge 0$.
3. Solve $(x - 2)^2(x + 1) > 0$.
4. Solve $\displaystyle \frac{x + 3}{x - 4} \le 0$.
5. Solve $\displaystyle \frac{x - 5}{x + 1} > 0$.
6. Solve $(3 - x)(x - 1) \ge 0$.
7. Solve $x^3 - x^2 - 6x < 0$.
8. Solve $\displaystyle \frac{x + 1}{x - 2} \ge 2$.
9. Solve $(x^2 + 4)(x + 2)(x - 1) \le 0$.
10. Solve $(x - 1)(x - 2)^2(x - 3)^3 \ge 0$.

Answers#