Video Transcript
Determine the solution set of the inequality π of π₯ is less than or equal to zero using the graph below.
What is the graphical interpretation of this inequality? Well, π of π₯ is less than or equal to zero when its graph is below or on the π₯-axis. Looking at the graph, we can see that there are two regions where this happens. One of them is to the left of negative four. And as weβre not given a circle to tell us the graph stops, we assume it continues all the way left and down. The second region is to the right of negative three. And again, we follow it all the way down assuming it continues on.
Weβre looking for the values of π₯ for which π of π₯ is less than or equal to zero. So for the orange part of the graph to the left of negative four, that region is given by π₯ is less than or equal to negative four. Weβre using the less than or equal to sign rather than just the less than sign because negative four is in this region. The endpoint is included. We can see from the graph that π of negative four is equal to zero, and so it satisfies π of π₯ is less than or equal to zero.
The other region is π₯ is greater than or equal to negative three where weβre using the greater than or equal to sign for the same reason as before because negative three is included in this region because π of negative three is zero and satisfies π of π₯ is less than or equal to zero. So essentially our solution set contains the values of π₯ for which either π₯ is less than or equal to negative four or π₯ is greater than or equal to negative three. We can also write this in interval notation. π₯ is less than or equal to negative four can be written as: open parentheses, negative infinity, negative four, close square brackets. And the fact that weβre using open parenthesis on the left tells us that the end point negative infinity is not included. But the fact that weβre using a close square brackets on the right tells us that the end point negative four is included.
Similarly, the inequality π₯ is greater than or equal to negative three can be written in interval notation as open square brackets, negative three, infinity, close parenthesis, where the open square brackets on the left-hand side tells us that the end point negative three is included in the interval, and the close parenthesis on the right-hand side tells us that the end point infinity is not. And to express the fact that π₯ can be in either one of these intervals, we use the union symbol. Thatβs one way to write the solution set in interval notation, but there is another way. We can instead describe the solution set as being everything apart from the small green region between negative four and negative three. This green region is everything but negative four is less than π₯ is less than negative three.
In interval notation, this is everything but open parenthesis, negative four, negative three, close parenthesis where weβve been careful to use parentheses rather than square brackets and less than signs rather than less than or equal to signs because neither end point is included in this interval. Remember that this is the interval of the values of π₯ which do not satisfy the inequality and so are not in the solution set.
In our case, everything is the set of real numbers. And so we can write everything but this interval as the real numbers minus the interval. Just a quick note to say, that this can also be written as β with a slash then the interval that should be removed.