We previously learned that solving equations is often best accomplished by finding an equivalent equation of the form \(A=0\). Then, by factoring the non-zero side, we are able to simplify the problem into testing where each factor individually equals zero. A similar strategy works for inequalities. However, there is a big difference — we can not work with only one factor at a time, or at least not quite.
The rule for equality with zero says \(AB=0\) if and only if \(A=0\) or \(B=0\). However, when the relation is changed to an inequality, there are multiple possibilities. For example, the inequality \(AB > 0\) says that the product \(AB\) is a positive number. This could happen if \(A\) and \(B\) are both positive, or it could happen if \(A\) and \(B\) are both negative. So if we were to solve \(A > 0\) and \(B > 0\), we are going to completely miss the other possibility.
However, suppose that we did not only solve \(A > 0\) but we also solved \(A < 0\).
(The other possibilities are \(A = 0\) or \(A\) is not defined.)
We can summarize our results using a number line divided into the relevant intervals.
For example, suppose we found \(A > 0\) on intervals \((0,4)\) and \(A < 0\) on
intervals \((-\infty,0)\) and \((4,\infty)\). The corresponding sign analysis
number line summary would look like the following.
If we create sign analysis summary number lines for both \(A\) and \(B\), then the simple rules for signs of products can be applied. (In mathematics, these are called parity rules and are related to even and odd numbers.) \[ \begin{align*} + \cdot + &= + \\ + \cdot - &= - \\ - \cdot + &= - \\ - \cdot - &= + \end{align*} \] We can then create a joint sign analysis number line for the product by finding all of the relevant intersections of the two number lines and show the resulting sign for the product.
Example: Use sign analysis to find the signs of \((x-3)(x+1)\).
The first factor, \(A=x-3\), is positive, \(x-3 > 0\), when \(x > 3\) and negative,
\(x-3 < 0\), when \(x < 3\). The sign analysis number line for \(x-3\) is given by
The second factor, \(B=x+1\), is positive when \(x > -1\) and negative when \(x < -1\).
The sign analysis number line for \(x+1\) is given by
When we look at overlapping intervals (intersection), we see three different
intervals: \((-\infty,-1)\) (where the signs are \(- \cdot -\)),
\((-1,3)\) (where the signs are \(- \cdot +\)),
and \((3,\infty)\) (where the signs are \(+ \cdot +\)).
From this, we construct the sign analysis number line summary for the product
\((x-3)(x+1)\) below.
We can interpret the sign analysis summary as giving answers to inequalities. The inequality \((x-3)(x+1) > 0\) is asking where the product is positive. So we interpret our sign analysis summary as giving the solution set, \[\{x : (x-3)(x+1) > 0 \} = (-\infty,-1) \cup (3,\infty). \] Similarly, the inequality \((x-3)(x+1) < 0\) is asking where the product is negative. Our sign analysis summary provides the solution set, \[\{x : (x-3)(x+1) < 0 \} = (-1,3). \]
Ordinarily, inequalities do not start in an obvious sign analysis form. In order to use this technique, you must first transform the inequality to a new inequality that relates an expression to zero. Then you must rewrite the expression in a factored form that is amenable to sign analysis. For polynomials, we want elementary factors, similar to \(x-3\) or \(x+1\), for which it is obvious how to find the signs.
Once we find the factorization, we could create a number line for each factor and then determine all of the relevant overlapping intervals. However, it is usually more direct to simply identify the points where each factor changes sign without actually creating the sign analysis summaries. Then, you create a single number line that includes each of the points where individual factors change sign. The resulting subintervals separated by these points will be exactly the relevant overlapping intervals.
After the intervals are identified, you can quickly determine the sign of the full expression by working through the signs of each factor. A slightly slower, but easier to remember strategy is to choose actual values in each subinterval and use those numbers to determine the sign of the overall product. By identifying which factor changes sign at each point, once the sign in one subinterval is determined, the remaining subintervals can be quickly determined.
Example: Solve the inequality \(x^2 \ge x+2\). The graph below should suggest the form of the solution set.
We start by rewriting the inequality as a relation with zero, since addition is an order-preserving operation \[ x^2-x-2 \ge 0. \] Next we rewrite our expression in factored form, \[ (x-2)(x+1) \ge 0. \] The factor \(x-2\) equals zero at \(x=2\), which also corresponds to the point where the factor changes sign. Similarly, the factor \(x+1\) changes sign at \(x=-1\), where the factor equals zero.
We create the sign analysis summary using the two points \(x=-1\) and \(x=2\). Because
a factor is zero at each point, we will also include a notation that the product equals
zero at each point. The blank sign analysis summary — blank because we will step
through the process of adding the sign information — is shown below.
We start in the furthest left interval, \((-\infty,-1)\). It is useful to choose
a single value in the interval, say \(x=-2\). The first factor, \(x-2\), would have
value \(x-2=-4\) which we summarize its sign as \(-\). The second factor, \(x+1\),
would have value \(x+1=-1\), which also has sign \(-\). We update the number line
summary with this information.
The point \(x=-1\) is where the second factor \(x+1\) changes sign. So we can immediately
update our number line with the signs of the two factors.
The point \(x=2\) is where the first factor \(x-2\) changes sign, and we again
update our number line.
Knowing the signs of the factors, we can use the parity rules to find the sign
of the entire product.
Once we have a sign analysis number line summary, we can find the solution set to the inequality. The formula \(x^2-x-2 = (x-2)(x+1)\) is positive on intervals \((-\infty,-1)\) and \((2,\infty)\), is negative on the interval \((-1,2)\), and is zero at the points \(\{-1,2\}\). Our question was transformed to \(x^2-x-2 \ge 0\), which is the union of the positive and zero solution sets to give \[ \{x : x^2 \ge x-2\} = (-\infty,-1] \cup [2,\infty). \]
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