About this video
Using a table of values for a polynomial h, the video identifies exact zeros where the table shows h(x) = 0 (x = -5 and x = 1). It then explains that because polynomials are continuous, a sign change between consecutive table values (positive at x = -1, negative at x = 0) guarantees another zero between them.
You'll be able to
- read exact zeros of a function from a table.
- explain why a continuous function that changes sign must cross zero.
- use a sign change in a table to locate an interval containing a zero.
Key words
- zero (solution of h(x) = 0)
- An x-value where the function's output is 0.
- continuous function
- A function whose graph has no jumps or breaks; all polynomials are continuous.
- sign change
- When the output goes from positive to negative, or negative to positive.
- consecutive x-values
- Two x-values that sit next to each other in the table.
Transcript
- [Instructor] So we're told that h of x is equal to all of this business over here, and they say "the table for h is shown below." So they give us some x values and what the corresponding h of that x value is. So h of -5 is 0, h of -4 is 25, and so forth. "Identify the exact x values where h of x is equal to 0."
So pause this video and have a go at that. Alright, so when does h of x equals 0? So we could see it equals 0 right over here and right over there. So what are the x values when h of x equals 0? Well, it's x equals -5, and it is x equals 1. Another way could say it, h of -5 is equal to 0, and h of 1 is equal to 0.
Alright, let's do the second part. "Between which two consecutive x values from the table must another solution to h of x equals 0 exists?" So pause this video and have another go at it. Alright, so one way to think about it is, in order for... It doesn't always have to be the case that...
Let's say this is a point right over here, this is some x value, let's call it x1, where we have some curve. I don't what h actually looks like. So you could have some point right over here where h of x1 is equal to 0, where before that point, you're going to have a positive value of h, and after that point, you have a negative value of h.
Now, if you were to look at this, let's call this x0, and you would see that h of x0 is greater than 0. h of x0 is greater than 0. And then this, let's call this x2, and then you'd say h of x2 is less than 0. As long as you know that this is a continuous function, you know that because this is just a traditional polynomial, you're not going to have any weird jumps or anything like that.
You know, hey, if we had to go from a positive value for h to a negative value of h, and it's continuous, we had to cross h of x equals 0. We had to cross the x intercept at... Or we had to cross the x axis, I should say, at some point in between. Now, you don't have to always have that for 0.
You could have a situation where you do something right like that. But if you have between two consecutive points that a continuous function is positive, and then that continuous function becomes negative, or vice versa, then you must have crossed 0. So let's think about where we have a change in sign.
We have 0 to positive. That's not quite a change in sign, we're still non-negative. So we're positive, positive, positive, and then negative. So we went from a positive h of x to a negative h of x. A positive h of x to a negative h of x. And that happened between x equals -1 and x equals 0.
So between these two consecutive x values,
- 1 and 0 from the table, a solution to h of x equals 0 must exist. At some point in between here, we don't know exactly, in this example, our x1 would sit right in between there, there's some x1 where h of x1 is equal to 0 because we had to go from positive to negative.
If it was also the other way around, from negative to positive, and it's a continuous function, we'd also be able to make that same conclusion.
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