It’s Professor Dave, let’s graph inequalities.
We’ve learned a lot about graphing functions, and graphing inequalities will be extremely
similar, but there are a few additional points to discuss, so let’s go through those now.
First, let’s recall what an inequality is.
Rather than an equation, where the two sides are equivalent to one another, inequalities
state that one side is greater than the other, or less than the other, or sometimes greater
than or equal to, or less than or equal to.
So if X is greater than or equal to three, then X could be three, or any real number
greater than three, including 3.001, pi, seven halves, or a billion.
The first thing we need to do is learn how to solve inequalities that are a little more
involved than the simple ones we’ve seen before.
Take something like X squared plus six X is less than negative five.
To solve this, we get everything on one side so that the function is either greater than
zero or less than zero.
In this case, we add five, and we get this.
Then, we pretend that the less than sign is an equals sign, and we solve the equation.
In this case, that would require factoring this to get X plus one times X plus five,
which gives us negative one and negative five as the solutions.
Then, we can draw a number line, and mark these points.
Then, to see which sections of the number line satisfy the inequality, we just pick
a point from each section and see if it works.
In between, we could try negative two.
Negative two squared is four, plus negative twelve is negative eight, plus five is negative
three.
Negative three is less than zero, so this section is a solution to the inequality.
Now let’s move over to the right.
Zero is always an easy one to try, so let’s use that.
Zero plus zero plus five is five, which is not less than zero, so from negative one to
positive infinity, these values don’t work.
Trying the other side, let’s use negative six.
Thirty-six plus negative thirty six gives us zero, plus five is five, so none of these
values work.
The solution to the inequality is everything from negative five to negative one, exclusive,
which we denote by writing this parenthetical term.
This approach will work for trickier inequalities with many terms on both sides, we just combine
all the like terms on one side so that the other side is zero, and then solve.
Whatever the zeroes are, we then just test points to find out which intervals make the cut.
When polynomial inequalities contain two variables, like Y is greater than or equal to X plus
two, these become interesting to graph.
When we do this, the first step is to treat this inequality as though it were an equation,
so Y equals X plus two.
As we should know at this point, this will just be the line Y equals X, but shifted up
two, given this vertical shift term here.
Now, what the inequality says is that Y must be greater than or equal to this.
So for any input X, Y must be greater by two or more.
So when X is zero, Y must be two or more.
When X is one, Y must be three or more.
That means that every point on this line is a solution to the inequality, as is every
point in the XY plane that is above this line, since every point above the line has a Y value
that is greater than the corresponding Y value of the line.
We would denote this by shading in this section of the coordinate plane, implying that everything
above this line is a solution to the inequality.
We can do the same thing for Y is greater than X squared minus four, we just graph the
function, and shade in everything above it, or the section containing points with Y values
that are greater than the Y value for the point on the function directly below.
Now let’s look at rational inequalities.
As we could probably guess after learning how to graph rational functions, these are
just inequalities where one side is a rational function.
Take something like two X plus two over X plus three is greater than zero.
We can solve this by figuring out a few things about it.
First, the X intercept.
We get that by setting the function equal to zero.
We can then multiply both sides by the denominator, which makes it vanish, because when multiplied
by zero it becomes zero.
Subtract two, divide by two, and negative one must be the X intercept.
Then, we can find the vertical asymptote by setting the denominator equal to zero.
X equals negative three must be the asymptote.
Then we can graph the function by testing points near the asymptote, intercept, and
approaching either infinity.
Once we arrive at a sketch of the function, we go back to the inequality.
This asks when the function is greater than zero.
So the question is, in what intervals is the function above the line Y equals zero?
That would be from negative infinity up until negative three, then from negative three to
negative one the function is negative, and then beyond negative one, the function is
positive again, so we get negative infinity to negative three, and negative one to positive
infinity, both exclusive, so we use parentheses, and we put a U for union in between, meaning
these two intervals together comprise the solution of the inequality.
We will learn more about this kind of notation when we go over set theory a little bit later.
That’s pretty much all there is to solving and graphing polynomial and rational inequalities,
so let’s check comprehension.