The sign analysis of a function consists in determining the values of the variable for which the function is positive, negative, or zero.
In other words, given a function \(f\), we want to establish where the following hold:
\[ f(x)>0,\qquad f(x)=0,\qquad f(x)<0. \]
This procedure is fundamental in the study of equations, inequalities, polynomial functions, rational functions, and, more generally, in analyzing the graph of a function.
Table of Contents
- What it means to analyze the sign of a function
- Positivity set, negativity set, and zeros
- General method for sign analysis
- Sign analysis of a product
- Sign analysis of a rational function
- Zeros of even and odd multiplicity
- Factors that are always positive or always negative
- Complete worked example
- Common mistakes
What it means to analyze the sign of a function
Analyzing the sign of a function means determining the intervals of its domain on which the function is positive, negative, or zero.
From a geometric standpoint:
- \(f(x)>0\) means that the graph of the function lies above the \(x\)-axis;
- \(f(x)<0\) means that the graph of the function lies below the \(x\)-axis;
- \(f(x)=0\) means that the graph crosses or touches the \(x\)-axis.
The zeros of the function are therefore the points at which the graph meets the \(x\)-axis.
Positivity set, negativity set, and zeros
Let \(f\) be a function defined on a domain \(D_f\).
The positivity set is the set of all \(x\in D_f\) for which:
\[ f(x)>0. \]
The negativity set is the set of all \(x\in D_f\) for which:
\[ f(x)<0. \]
The zeros of the function are the values in the domain for which:
\[ f(x)=0. \]
It is important to stress that zeros must belong to the domain of the function. A value that makes a denominator vanish, for instance, is not a zero of the function: it is a point excluded from the domain.
General method for sign analysis
The general method for analyzing the sign of an algebraic function rests on a few key steps.
1. Determine the domain
The first step is to find the domain \(D_f\), that is, the set of all values for which the function is defined.
For example, if:
\[ f(x)=\frac{x-1}{x+3}, \]
the denominator must be nonzero. Therefore:
\[ x+3\neq 0, \]
which gives:
\[ x\neq -3. \]
The domain is:
\[ D_f=\mathbb{R}\setminus\{-3\}. \]
2. Factor the function
Whenever possible, the function should be written as a product of simpler factors.
For example:
\[ x^2-5x+6=(x-2)(x-3). \]
Factoring allows us to study the sign of each factor separately.
3. Find the zeros and excluded points
The zeros of the function are found by setting the numerator equal to zero or, in the case of a product, by setting each factor equal to zero in turn.
Excluded points are found by identifying the values that make the denominator zero or that otherwise render the function undefined.
4. Arrange the key values on the number line
The zeros and excluded points partition the real number line into intervals. On each such interval the sign of the function remains constant, provided the function is composed of continuous factors and does not vanish within the interval.
5. Build the sign chart
Finally, construct a sign chart by determining the sign of each factor on every interval, then combining the results.
Sign analysis of a product
Consider a function written as a product of factors:
\[ f(x)=A(x)\cdot B(x). \]
The sign of \(f(x)\) is determined by the signs of the two factors.
Recall the fundamental rules:
\[ (+)\cdot(+)=+,\qquad (-)\cdot(-)=+, \]
while:
\[ (+)\cdot(-)=-,\qquad (-)\cdot(+)=-. \]
Thus a product is positive when it contains an even number of negative factors, and negative when it contains an odd number of negative factors.
Example
Analyze the sign of:
\[ f(x)=(x+1)(x-3). \]
The zeros are:
\[ x+1=0 \Rightarrow x=-1, \qquad x-3=0 \Rightarrow x=3. \]
The points \(-1\) and \(3\) divide the real number line into the intervals:
\[ (-\infty,-1),\qquad (-1,3),\qquad (3,+\infty). \]
We study the sign of each factor:
\[ \begin{array}{c|ccc} x & (-\infty,-1) & (-1,3) & (3,+\infty)\\ \hline x+1 & - & + & +\\ x-3 & - & - & +\\ \hline f(x) & + & - & + \end{array} \]
Therefore:
\[ f(x)>0 \text{ for } x<-1 \text{ or } x>3, \]
\[ f(x)=0 \text{ for } x=-1 \text{ and } x=3, \]
\[ f(x)<0 \text{ for } -1
Sign analysis of a rational function
A rational function has the form:
\[ f(x)=\frac{A(x)}{B(x)}. \]
In this case it is essential to distinguish carefully between:
- the zeros of the numerator, which may be zeros of the function;
- the zeros of the denominator, which are points excluded from the domain.
A fraction is positive when numerator and denominator have the same sign:
\[ \frac{+}{+}=+,\qquad \frac{-}{-}=+. \]
It is negative when numerator and denominator have opposite signs:
\[ \frac{+}{-}=-,\qquad \frac{-}{+}=-. \]
Example
Analyze the sign of:
\[ f(x)=\frac{x-1}{x+3}. \]
The denominator vanishes at:
\[ x+3=0 \Rightarrow x=-3, \]
so \(x=-3\) is excluded from the domain.
The numerator vanishes at:
\[ x-1=0 \Rightarrow x=1. \]
We study the sign on the intervals determined by \(-3\) and \(1\):
\[ (-\infty,-3),\qquad (-3,1),\qquad (1,+\infty). \]
The sign chart is:
\[ \begin{array}{c|ccc} x & (-\infty,-3) & (-3,1) & (1,+\infty)\\ \hline x-1 & - & - & +\\ x+3 & - & + & +\\ \hline f(x) & + & - & + \end{array} \]
Therefore:
\[ f(x)>0 \text{ for } x<-3 \text{ or } x>1, \]
\[ f(x)=0 \text{ for } x=1, \]
\[ f(x)<0 \text{ for } -3
The point \(x=-3\) is not a zero: it is excluded from the domain.
Zeros of even and odd multiplicity
An important aspect of sign analysis concerns the multiplicity of zeros.
Consider a factor of the form:
\[ (x-a)^m. \]
The integer \(m\) is called the multiplicity of the zero \(x=a\).
Odd multiplicity
If \(m\) is odd, the factor changes sign as \(x\) passes through \(a\).
For example:
\[ (x-2)^3 \]
is negative for \(x<2\) and positive for \(x>2\):
\[ (x-2)^3<0 \text{ for } x<2, \]
while:
\[ (x-2)^3>0 \text{ for } x>2. \]
Even multiplicity
If \(m\) is even, the factor does not change sign as \(x\) passes through \(a\).
For example:
\[ (x-2)^2 \]
is always nonnegative:
\[ (x-2)^2\ge 0 \]
for every \(x\in\mathbb{R}\), and equals zero only at \(x=2\).
In particular:
\[ (x-2)^2>0 \text{ for } x\neq 2. \]
For this reason, a zero of even multiplicity does not produce a sign change in the sign chart.
Factors that are always positive or always negative
Certain factors never change sign.
For example:
\[ x^2+1>0 \]
for every \(x\in\mathbb{R}\), since \(x^2\ge 0\) implies that \(x^2+1\) is always strictly positive.
Likewise, a perfect square such as:
\[ (x-3)^2 \]
is always nonnegative:
\[ (x-3)^2\ge 0. \]
It vanishes only at \(x=3\), but does not change sign at that point.
Recognizing such factors is very useful, as it simplifies the sign analysis considerably.
Complete worked example
Analyze the sign of:
\[ f(x)=\frac{(x+1)^2(x-2)}{x^2(x+3)}. \]
Domain
The denominator is:
\[ x^2(x+3). \]
It vanishes at:
\[ x^2=0 \Rightarrow x=0, \]
or at:
\[ x+3=0 \Rightarrow x=-3. \]
Therefore:
\[ D_f=\mathbb{R}\setminus\{-3,0\}. \]
Zeros of the function
The zeros are found by setting the numerator equal to zero:
\[ (x+1)^2(x-2)=0. \]
This gives:
\[ x=-1,\qquad x=2. \]
The value \(x=-1\) is a zero of multiplicity two, since the factor \((x+1)^2\) appears in the numerator.
Sign analysis of each factor
The factors to consider are:
\[ (x+1)^2,\qquad x-2,\qquad x^2,\qquad x+3. \]
The factors \((x+1)^2\) and \(x^2\) are always nonnegative and do not change sign.
The sign of the function is therefore governed by the factors \(x-2\) and \(x+3\), taking into account the zeros and excluded points.
The key values are:
\[ -3,\qquad -1,\qquad 0,\qquad 2. \]
The sign chart is:
\[ \begin{array}{c|ccccc} x & (-\infty,-3) & (-3,-1) & (-1,0) & (0,2) & (2,+\infty)\\ \hline (x+1)^2 & + & + & + & + & +\\ x-2 & - & - & - & - & +\\ x^2 & + & + & + & + & +\\ x+3 & - & + & + & + & +\\ \hline f(x) & + & - & - & - & + \end{array} \]
Conclusion
The function is positive for:
\[ x<-3 \text{ or } x>2. \]
The function is negative for:
\[ -3
The function equals zero at:
\[ x=-1,\qquad x=2. \]
The points:
\[ x=-3,\qquad x=0 \]
are excluded from the domain.
Common mistakes in sign analysis
Overlooking the domain
For rational functions, determining the domain is the first step. Any value that makes the denominator vanish must be excluded, even if that factor is later cancelled.
Confusing zeros with excluded points
A zero of the function is a value for which \(f(x)=0\). An excluded point, by contrast, does not belong to the domain of the function at all.
For example, in:
\[ f(x)=\frac{x-1}{x+3}, \]
\(x=1\) is a zero, while \(x=-3\) is excluded from the domain.
Ignoring the multiplicity of zeros
A zero of odd multiplicity produces a sign change. A zero of even multiplicity does not.
Cancelling factors without preserving the exclusions
Consider:
\[ f(x)=\frac{(x-1)(x-2)}{(x+3)(x-2)}. \]
For \(x\neq 2\), the factor \(x-2\) may be cancelled:
\[ f(x)=\frac{x-1}{x+3}. \]
However, \(x=2\) remains excluded from the domain of the original function.
This is a crucial point: cancellation simplifies the expression, but it does not remove the existence conditions of the original function.
In summary, sign analysis is an essential tool for understanding the behavior of a function. The correct approach involves finding the domain, factoring the expression, identifying zeros and excluded points, analyzing the sign of each factor, and finally combining the results in a sign chart.