The sign of a function

Build a table of signs using test points around zeroes and discontinuities, and use it to solve polynomial and fraction inequations and to start a sketch.

Tables of signs

So find all the zeroes and discontinuities, then test one value in each gap. That is the whole method.

Example 1●●

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

Zeroes: $x = -2, 1, 3$. There are no discontinuities.

$x$	$-3$	$-2$	$0$	$1$	$2$	$3$	$4$
$y$	$-24$	$0$	$6$	$0$	$-4$	$0$	$18$
sign	$-$	$0$	$+$	$0$	$-$	$0$	$+$

Test one point in each gap, for example x = −3, 0, 2, 4.

$-2 \le x \le 1$ or $x \ge 3$

Positive or zero, so include the zeroes.

Fractions and discontinuities

A fraction can change sign where its denominator is zero, so put those $x$-values in the table too. The function is undefined there, so a discontinuity is never part of the solution.

Sign of y = (x + 2)(x − 1) / (x − c)

The shading shows where y is positive. The sign changes only at the zeroes and at the asymptote. Move c and watch the pattern.

Try this

Move $c$ between $-2$ and $1$. How does the solution of $y > 0$ change? What happens when $c = 1$?

Example 2●●

Solve $\dfrac{x + 1}{x - 2} \le 0$.

Zero at $x = -1$, discontinuity at $x = 2$.

At $x = -2$: $\tfrac{-1}{-4} > 0$. At $x = 0$: $\tfrac{1}{-2} < 0$. At $x = 3$: $\tfrac{4}{1} > 0$.

Test x = −2, 0 and 3.

$-1 \le x < 2$

Include the zero, exclude the discontinuity.

Comparing with the method of 1A

The fraction inequations from the last section can also be done with a table of signs. Move everything to one side and combine into one fraction first.

Example 3●●●

Solve $\dfrac{5}{x - 4} \ge 1$ using a table of signs.

$\dfrac{5}{x - 4} - 1 \ge 0$, so $\dfrac{5 - (x - 4)}{x - 4} = \dfrac{9 - x}{x - 4} \ge 0$.

Signs: $-$ for $x < 4$, $+$ for $4 < x < 9$, $-$ for $x > 9$.

Zero at 9, discontinuity at 4. Test x = 0, 5 and 10.

$4 < x \le 9$, the same answer as before.

Flashcards

Where can a function change sign?

Only at its zeroes and its discontinuities.

Is a discontinuity ever included?

No. The function is undefined there.

Double zero

The graph touches the axis and the sign does not change.

Practice

A function can only change sign at:
Evaluate $(x + 2)(x - 1)(x - 3)$ at the test point $x = 0$.
Solve $(x + 2)(x - 1)(x - 3) \ge 0$.
Solve $\dfrac{x + 1}{x - 2} \le 0$.
How many integers satisfy $\dfrac{x + 1}{x - 2} \le 0$?
Solve $(x - 1)^2(x + 2) > 0$.
Solve $x^3 > 4x$.
At what $x$-value is the discontinuity of $y = \dfrac{x(x - 4)}{x + 3}$?
$\dfrac{5}{x - 4} - 1$ written as one fraction is: