Introduction and the Importance of Zeros

In the previous section, we constructed the following graph to solve the inequality , where .

We can see that the graph of dips below the -axis between its two -intercepts. The zeros of are and in this case and they divide the domain (the -axis) into three intervals: , and . For every number in , the graph of is above the -axis; in other words, for all in . Similarly, for all in , and for all in . We can schematically represent this with the sign diagram below.

Here, the above a portion of the number line indicates for those values of ; the indicates there. The numbers labeled on the number line are the zeros of , so we place above them. We see at once that the solution to is . Adding in the zeros, the solution to is .

Our next goal is to establish a procedure by which we can generate the sign diagram without graphing the function.

Continuity and the Intermediate Value Theorem

An important property of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphically, this means that a parabola can’t be above the -axis at one point and below the -axis at another point without crossing the -axis.

This is a special case of a theorem called the Intermediate Value Theorem, or IVT for short. To talk about the IVT, we first need to discuss what it means for a function to be continuous.

In further courses, you will learn a more formal definition of continuity, but for now, this will suffice.

The function whose graph is shown as the first of the two graphs above is continuous, while the function whose graph is shown second above is not, since it has breaks in its graph.

One way to think about continuous functions is that they are the functions whose graphs you could draw on an infinite piece of paper without ever taking your pencil off the paper (except where they aren’t defined). You will encounter and learn about continuous functions more in-depth in calculus, but for now, familiarity at this level will be enough.

Now that we know about continuous functions, we can state our version of the IVT.

Reinterpreted, this means that the graph of a continuous function can’t be above the -axis at one point and below the -axis at another point without crossing the -axis.

Here’s how we’ll use the IVT to solve inequalities of the form , where is a continuous function. If a given interval does not contain a zero of , then by the IVT either all the function values on the interval are positive or they’re all negative. In this way, the IVT allows us to determine the sign of all of the function values on the interval by testing the function at just one value in the interval, which we’re free to choose.

This gives us the following steps for solving an inequality involving a continuous function.

(a)
Rewrite the inequality, if necessary, as a continuous function on one side of the inequality and on the other.
(b)
Find the zeros of and place them on the number line with the number above them.
(c)
Choose a real number, called a test value, in each of the intervals determined in step 2.
(d)
Determine the sign of for each test value in step c, and write that sign above the corresponding interval.
(e)
Choose the intervals which correspond to the correct sign to solve the inequality.

As you can see, the zeros of continuous functions are important, so in the examples that follow, we’ll highlight the techniques we use to find zeros. It may also be useful to review methods for finding zeros that you’ve seen before.

Solving Inequalities Algebraically

Dealing with Difficult Denominators

Even after we feel comfortable with the procedure for solving inequalities involving continuous functions, you might still wonder about functions which aren’t defined on all real numbers, such as rational functions or more generally, functions with denominators that could potentially evaluate to 0. The good news is that if and are continuous functions, the function is continuous wherever it is defined. Therefore, we can adapt our technique from before, but remembering that a change of sign could happen around a point where a function is undefined, so we need to add any places our functions are undefined to our sign diagram.

Viewing the left hand side as a rational function we make a sign diagram. The candidates for zeros of are the solutions to , which we can find by factoring. Therefore, the candidates for zeros of are , and . However, is not in the domain of , since it is the solution to , which is the equation we get by setting the denominator equal to 0. However, -values for which the function is undefined are also possible places where the sign of the function might change, so we should include them on the sign diagram. Since is a rational function, it is continuous everywhere it is defined, so when constructing the sign diagram, we only need to consider the intervals between zeros or places where it is undefined. For us, these intervals will be , , , and .

Choosing test values in each test interval (we encourage you to check the calculation), we can construct the sign diagram below.

We used an X to denote that is not defined at .

We are interested in where . We find is positive on the intervals , and . We add to these intervals the zeros of , , and , to get our final solution: .

We set . We note that is undefined when its denominator , or when . Solving this by taking logarithms gives , so the domain of is . To find the zeros of , we set the numerator equal to zero and obtain . Solving for , we find , or . When we build our sign diagram, finding test values may be a little tricky since we need to check values around and . Recall that the function is increasing1 which means . This indicates that we might want to use , , and as our test values. While the prospect of determining the sign of may be very unsettling, remember that , so We determine the signs of and similarly and construct the sign diagram.

From the sign diagram, we find our answer to be .

Conclusion

We hope that the specific examples we’ve gone through illustrate a general principle when it comes to solving inequalities. First, we want to rewrite the inequality in a form where 0 is on one side and a nice-enough function is on the other side. Then, we use the fact that our functions are continuous on their domains to narrow down where possible sign changes can occur. From there, we can use test values to compute the sign of the function on intervals, and finish by putting our solution in interval notation.