The Sign of a Function Revision Notes for HSC SSCE Mathematics Advanced

The Sign of a Function

Understanding the sign of a function

When we sketch a curve or solve inequations, we need to know where the function is above the $x$ -axis (positive) and where it is below the $x$ -axis (negative). The sign of a function tells us exactly this information.

For any function $f(x)$ , we can classify every value in its domain as one of three types:

Positive: where $f(x) > 0$ (the curve is above the $x$ -axis)
Negative: where $f(x) < 0$ (the curve is below the $x$ -axis)
Zero: where $f(x) = 0$ (the curve crosses or touches the $x$ -axis)

infoNote

Understanding the sign of a function is essential for solving inequations and analyzing function behavior. Any problem involving "greater than" or "less than" can be transformed into a question about where a function is positive or negative.

This becomes especially important when solving inequations. Any inequation can be rearranged so all terms are on one side. For example:

$x^3 + 1 \geq x^2 + x$

can be rewritten as:

$x^3 - x^2 - x + 1 \geq 0$

This means solving an inequation is the same as finding where a function is positive, negative, or zero.

Bracket interval notation

What is bracket interval notation?

Bracket interval notation is an alternative way to write intervals. Instead of using inequalities like $1 \leq x \leq 3$ , we can write $[1, 3]$ . This notation is more concise and particularly useful when working with unions of intervals.

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The key rule is:

Square bracket [ or ] means the endpoint is included (like $\leq$ or $\geq$ )
Round bracket ( or ) means the endpoint is not included (like $<$ or $>$ )

Think of it this way: a square bracket looks like a solid wall that holds the endpoint in, while a round bracket curves away from the endpoint, excluding it.

Types of intervals

Here are the main types of intervals written in both notations:

Diagram	Using inequalities	Using brackets
Solid dots at $\frac{1}{3}$ and $3$	$\frac{1}{3} \leq x \leq 3$	$[\frac{1}{3}, 3]$
Open dots at $-1$ and $5$	$-1 < x < 5$	$(-1, 5)$
Solid dot at $-2$ , open dot at $3$	$-2 \leq x < 3$	$[-2, 3)$
Solid dot at $-5$ , arrow right	$x \geq -5$	$[-5, \infty)$
Arrow left, open dot at $2$	$x < 2$	$(-\infty, 2)$

Closed intervals

A closed interval like $[a, b]$ contains both its endpoints. For example, $[1, 3]$ includes $1$ , $3$ , and all numbers in between.

Open intervals

An open interval like $(a, b)$ does not contain either endpoint. For example, $(-1, 5)$ includes all numbers between $-1$ and $5$ , but not $-1$ or $5$ themselves.

Mixed intervals

Intervals like $[-2, 3)$ or $(a, b]$ contain one endpoint but not the other.

Unbounded intervals

When an interval extends forever in one direction, we use the infinity symbol $\infty$ :

$[a, \infty)$ means all numbers greater than or equal to $a$
$(a, \infty)$ means all numbers greater than $a$
$(-\infty, a]$ means all numbers less than or equal to $a$
$(-\infty, a)$ means all numbers less than $a$

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The symbol $\infty$ (infinity) is not a number—it's a concept indicating that an interval continues without bound. This is why we always use a round bracket with $\infty$ or $-\infty$ : infinity itself can never be included as an endpoint.

Special cases

The notation $(-\infty, \infty)$ represents all real numbers
The notation $[a, a]$ represents the single point $\{a\}$ (called a degenerate interval)

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Common Confusion Alert!

Be careful: the notation $(2, 5)$ could mean either an open interval OR a point in the coordinate plane, depending on context. Always check the context to determine which interpretation is correct.

The union of intervals

Sometimes the solution to an inequation consists of multiple separate intervals. We use the union symbol $\cup$ to combine them.

For example, consider the function $y = x(x - 3)$ . This is a parabola that crosses the $x$ -axis at $x = 0$ and $x = 3$ .

The inequation $x(x - 3) \geq 0$ has solution:

$x \leq 0 \text{ or } x \geq 3$

In bracket notation, this is written as:

$(-\infty, 0] \cup [3, \infty)$

Here are more examples showing the connection between "or" and the union symbol:

Diagram	Using inequalities	Using brackets
Solid intervals $[0,1]$ and $[2,3]$	$0 \leq x \leq 1$ or $2 \leq x \leq 3$	$[0, 1] \cup [2, 3]$
Mixed dots at points	$-1 < x \leq 1$ or $3 \leq x < 6$	$(-1, 1] \cup [3, 6)$
Left ray and open interval	$x \leq 2$ or $3 < x < 4$	$(-\infty, 2] \cup (3, 4)$

infoNote

The word "or" in mathematics corresponds directly to the union symbol $\cup$ in set notation. When you see "or" between two conditions, you can translate it to $\cup$ in bracket notation.

Where can a function change sign?

Understanding where a function can change from positive to negative (or vice versa) is crucial for solving inequations.

Zeroes

A zero of a function is a value where $f(x) = 0$ . On a graph, this is where the curve crosses or touches the $x$ -axis. A function may change sign at a zero.

Discontinuities

A discontinuity is a point where the function has a break or gap. Informally, if you can't draw the curve through a point without lifting your pen, that point is a discontinuity. A function may also change sign at a discontinuity.

The key principle

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Fundamental Principle of Sign Changes:

A function can only change sign at zeroes or discontinuities.

Between any two consecutive zeroes and discontinuities, the function must maintain the same sign. This is because if a function is continuous on an interval and never equals zero in that interval, it cannot cross the $x$ -axis, so it must stay entirely above or entirely below the axis.

In the graph above, notice:

The function has zeroes at $x = a$ and $x = b$
It has discontinuities at $x = c$ , $x = d$ , $x = e$ , and $x = f$
The function changes sign at zero $x = a$ and at discontinuities $x = c$ and $x = e$
It does NOT change sign at zero $x = b$ or at discontinuities $x = d$ and $x = f$

infoNote

Just because a function has a zero or discontinuity doesn't mean it must change sign there—but these are the only places where a sign change is possible. This is why identifying all zeroes and discontinuities is the first step in analyzing function sign.

Using a table of signs

A table of signs is a systematic method to determine where a function is positive and where it is negative.

How to create a table of signs

Step 1: Find all zeroes by solving $f(x) = 0$

Step 2: Find all discontinuities (usually where a denominator equals zero)

Step 3: Choose test values between consecutive zeroes and discontinuities

Step 4: Calculate $f(x)$ at each test value

Step 5: Record the sign (positive or negative) in each interval

The table structure

A table of signs has three rows:

Row 1: $x$ -values (the test points)
Row 2: $y$ -values (the function values at those points)
Row 3: sign ( $+$ for positive, $-$ for negative, $0$ for zero, $*$ for discontinuity)

infoNote

The sign remains constant between consecutive critical points (zeroes and discontinuities), so one test value per interval is sufficient. You don't need to test multiple points in the same interval.

Worked example: Polynomial inequation

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Worked Example: Polynomial Inequation

Consider the function $y = (x - 1)(x - 3)(x - 5)$

Step 1: Find the zeroes

Setting $y = 0$ : $(x - 1)(x - 3)(x - 5) = 0$

So $x = 1$ , $x = 3$ , or $x = 5$

Step 2: Check for discontinuities

This is a polynomial, so there are no discontinuities.

Step 3: Create the table of signs

We need test values in the intervals: $(-\infty, 1)$ , $(1, 3)$ , $(3, 5)$ , and $(5, \infty)$

Good choices are: $x = 0, 2, 4, 6$

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$	$-15$	$0$	$3$	$0$	$-3$	$0$	$15$
sign	$-$	$0$	$+$	$0$	$-$	$0$	$+$

From the table:

$y$ is positive for $1 < x < 3$ or $x > 5$ , written as $(1, 3) \cup (5, \infty)$
$y$ is negative for $x < 1$ or $3 < x < 5$ , written as $(-\infty, 1) \cup (3, 5)$

Solving the inequation $(x - 1)(x - 3)(x - 5) \leq 0$ :

We need where the function is negative or zero, so:

$x \leq 1 \text{ or } 3 \leq x \leq 5$

In bracket notation: $(-\infty, 1) \cup (3, 5)$

Worked example: Polynomial with repeated factor

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Worked Example: Polynomial with Repeated Factor

Solve $x^3 + 1 \leq x^2 + x$

Step 1: Rearrange with all terms on the left

$x^3 - x^2 - x + 1 \leq 0$

Step 2: Factor the expression

$y = x^3 - x^2 - x + 1$

$= x^2(x - 1) - (x - 1)$

$= (x^2 - 1)(x - 1)$

$= (x + 1)(x - 1)^2$

Step 3: Find zeroes

The zeroes are $x = -1$ and $x = 1$ (with $x = 1$ being a repeated zero)

Step 4: Create the table of signs

$x$	$-2$	$-1$	$0$	$1$	$2$
$y$	$-9$	$0$	$1$	$0$	$3$
sign	$-$	$0$	$+$	$0$	$+$

Step 5: Sketch and solve

The function is negative or zero when $x \leq -1$ or $x = 1$

In bracket notation: $(-\infty, -1) \cup (1, 1)$

Note: At the repeated zero $x = 1$ , the function touches the $x$ -axis but doesn't cross it, so the sign doesn't change.

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Understanding Repeated Zeroes:

When a zero appears with an even power (like $(x - 1)^2$ ), the function will touch the $x$ -axis but not cross it. The sign stays the same on both sides. When a zero appears with an odd power, the function crosses the $x$ -axis and changes sign.

Worked example: Rational function with discontinuity

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Worked Example: Rational Function with Discontinuity

Examine the sign of $y = \frac{x - 1}{x - 4}$

Step 1: Find zeroes

The numerator is zero when $x - 1 = 0$ , so $x = 1$

Step 2: Find discontinuities

The denominator is zero when $x - 4 = 0$ , so there's a discontinuity at $x = 4$

Step 3: Create the table of signs

We need test values in the intervals: $(-\infty, 1)$ , $(1, 4)$ , and $(4, \infty)$

$x$	$0$	$1$	$2$	$4$	$5$
$y$	$\frac{1}{4}$	$0$	$-\frac{1}{2}$	$*$	$4$
sign	$+$	$0$	$-$	$*$	$+$

The asterisk ( $*$ ) indicates the function is undefined at $x = 4$ .

From the table:

$y$ is positive for $x < 1$ or $x > 4$
$y$ is negative for $1 < x < 4$

This rational function has both a vertical asymptote (at $x = 4$ ) and a horizontal asymptote, which you'll study in the next section.

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For rational functions, always check both the numerator (for zeroes) and the denominator (for discontinuities). The discontinuities are often where vertical asymptotes occur.

Worked example: Always positive function

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Worked Example: Always Positive Function

Examine the sign of $y = \frac{1}{1 + x^2}$

Analysis:

The numerator is $1$ , which is never zero, so there are no zeroes.

The denominator is $1 + x^2$ . Since $x^2 \geq 0$ for all real $x$ , we have $1 + x^2 \geq 1 > 0$ for all real $x$ . So there are no discontinuities.

Conclusion: The function is always positive.

We can verify this with a single test value:

$x$	$0$
$y$	$1$
sign	$+$

Since there are no zeroes or discontinuities where the sign could change, and we've found the function is positive at one point, it must be positive everywhere.

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When a function has no zeroes or discontinuities, it maintains the same sign everywhere in its domain. In such cases, you only need to test a single value to determine the sign for all real numbers.

Remember!

bookmarkSummary

Key Points to Remember:

Bracket interval notation uses square brackets [ ] for included endpoints and round brackets ( ) for excluded endpoints. It provides a concise way to write intervals.
A function can only change sign at zeroes or discontinuities. Between these critical points, the sign remains constant.
A table of signs is created by choosing test values between zeroes and discontinuities. This systematic method reveals where the function is positive, negative, or zero.
To solve an inequation, rearrange it so all terms are on one side, then use a table of signs to find where the function has the required sign.
For polynomial inequations, you only need to consider zeroes. For rational functions, you must also consider discontinuities (where the denominator is zero).
The word "or" in mathematics corresponds to the union symbol $\cup$ when combining intervals.

The Sign of a Function (HSC SSCE Mathematics Advanced): Revision Notes