Solving rational inequalities from a graph
At the end the narrator says a less-than-or-equal version would include x = -2, but omits that x = 2 would also be included. A note covers both.
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Using the graph of g(x) = (x^2 - 4)/(x^2 - 9), the video finds where g(x) < 0. The graph is below the x-axis between the vertical asymptote x = -3 and the zero x = -2, and between the zero x = 2 and the asymptote x = 3, giving (-3, -2) ∪ (2, 3).
You'll be able to
- read from a graph the intervals where a function is less than zero.
- explain why vertical asymptotes are never included in a solution set.
- write a solution in inequality form and in interval notation with a union.
Key words
- Vertical asymptote
- A vertical line x = a that the graph approaches but never reaches, because the function is undefined there.
- Interval notation
- A way to write a range of x-values, using round brackets ( ) when the endpoints are not included.
- Union (∪)
- Combines two or more intervals into one solution set; read it as 'or'.
Transcript
- [Instructor] We're told the g of x is equal to x-squared minus four over x-squared minus nine. Use the graph to identify when g of x is less than zero. So pause this video, have a go at that. All right, we see here this is the graph of y is equal to g of x and we can see clearly when it is less than zero.
I will highlight when it's less than zero. This whole period, it is less than zero and it looks like it never quite... Yeah, it looks like we... Yeah, we definitely have a vertical asymptote right over here at x equals negative three and x equals positive three.
So it looks like we're going from, we're going from not quite, well, less than zero, so we're not even saying less than or equal to zero. So we're going from x needs to be greater than negative three and less than negative two. That's that part right over there.
So negative three is less than x is less than negative two. And then we go to this part right over here where two is less than x, which is less than three. That's this one right over here, which is less than three. Or another way we could write it. We could write it, we could go from the interval, not including from negative three to negative two.
And the union of this interval and the interval not including the edges, not including the boundaries from two to three. And we're done, and once again, we can see that visually. We just have to be very conscientious of these vertical asymptotes to recognize that we're never getting quite to x equals three there.
And it's very clear this was less than zero. If it was less than or equal to zero, then we would've included x equals negative two in our answer.
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