Algebra (all content)
- Introduction to average rate of change
- Worked example: average rate of change from graph
- Worked example: average rate of change from table
- Average rate of change: graphs & tables
- Worked example: average rate of change from equation
- Average rate of change of polynomials
Finding the interval where a function has an average rate of change of ½ given its equation. Created by Sal Khan.
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- Would the interval include -2 and 2?(40 votes)
- That's a good catch, Brian--the interval "-2<x<2" does not include the numbers -2 & 2 (it's what's called an open interval). However, the average value of the function is the same regardless.
Also, it's good to keep this in mind because this topic is especially important later on when you learn Calculus and the all-important Mean Value Theorem. Anyway, hope this helps!(41 votes)
- The video narrator just puts in −2 and 2 into the function to calculate the change.
But why? Isn't it an open interval? Aren't −2 and 2 outside the interval?(5 votes)
- Technically yes. However, for the purposes of computing the average, you can treat the otherwise open intervals as closed.(6 votes)
- How can we optimize our time in answering these type of questions (objective) without having to try out all the other options? (say, A, B, C, or D)(6 votes)
- There's no real way to do it besides trying out all the options. If you want to try and be faster you could look over the options and see if any of them look more likely and try that one first.
If you have a calculator that can use variables you may want to first work out f(x) for all the different end points of the intervals, keeping the function on your calculator, and then after this go back and calculate the slopes.(4 votes)
- Is there a way not to guess the numbers to solve for the possible value range for x if the average rate of change is an integer?(3 votes)
- Will it work to build the following equation?
1/8*x^3 = 1/2x (x= +-2 therefore between the points -2 and 2 the slope will be 1/2)
I was thinking, since the average rate of change equals to the slope of the function, if I find where the function
s slope is equal to 1/2x I could match it to the respective interval.s not a linear functions with all these exponents, and I`m not sure if adressing to the 1/8x^3 as the slope in this formation is correct?
The thing is that it
If this way is incorrect, is there any way of figuring the interval a function has a specific average slope(1/2x in this example) without being given specific interval?
I hope my question is clear.(2 votes)
- I have viewed all the videos from grade 3 through grade 8, math & pre-algebra, and algebra to this point; and I don't recall a formal introduction to "intervals." I do recall seeing inequalities using this form, but not learning to call them intervals or any specific properties relating to intervals. Is the use of the term interval in this video simply a way to provide x-values to plug into the given equations, or have I missed some prior introduction to intervals?(2 votes)
- In this context, an interval is just a range with a beginning and an end. So an interval of x defines a starting point for x and a stopping point for x. This video deals with intervals start about 3 minutes in: https://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/compound-inequalities
There don't appear to be any videos specifically around intervals and interval notation. This would be a good thing to request at https://khanacademy.wufoo.com/forms/suggest-a-topic(1 vote)
- This is a different equation, it's not the classic y = mx + b; it's y = mx + x. We can't say that m denotes slope, or rate of change for y, in this second equation, right? Because x is variable, it always changes the change of y. Can someone please explain this?(1 vote)
- The function Sal is using is not a linear function, so it should not be y = mx + b. However, the average rate of change is the slope of the line that connects two points. So you can import some of your ideas about how lines work in order to calculate that slope.(2 votes)
- is saying -2<x<2 is the same as saying (-2, 2), correct? or am I just crazy?(1 vote)
- Yes, -2<x<2 is the same as (-2,2). If it were -2<= x <=2, it would be brackets around it [ -2,2 ]
You can also have a parenthesis and a bracket like (-2,2]. It means that -2 is not included in the function but 2 is.(3 votes)
y equals 1/8 x to the third minus x squared. Over which interval does y of x have an average rate of change of 1/2? So let's go interval by interval and calculate the average rate of change. So first let's think about this interval right over here. x is between negative 2 and 2. So negative 2 is less than x, which is less than 2. So let's just think about what is the value of our function when x is equal to negative 2? So y of negative 2 is equal to 1/8 times negative 2 to the third power minus negative 2 squared, which is equal to-- let's see, negative 2 the third power is negative 8. Negative 8 divided by 8 is negative 1. Negative 2 squared is positive 4, but then you're going to have to subtract that, so it's minus 4. So this is equal to negative 5. And y of 2 is equal to 1/8 times 2 to the third power minus 2 squared. And that's going to be equal to 1/8 times 8 is 1 minus 4, which is equal to negative 3. So if you want to find your average rate of change, you want to figure out how much does the value of your function change, and divide that by how much your x has changed. So we could make a table here. x, y. When x is negative 2, y is negative 5. When x is positive 2, y is negative 3. So how much did your y change? Well, your y increased by 2, and your x increased by 4. And you could get these numbers-- you could just look at it that y increased from that point to that point, x increased from that point to that point. Or you could say hey, negative 3 minus negative 5 is positive 2. That's the difference between negative 3 and negative 5. If you said 2 minus negative 4, well that gives you, once again, the distance, or the difference. It would give you positive 4. But if you look at it here, it's clear. When y increased by 2-- Or we could say when x increased by 4, y increased by 2. So our average rate of change over this interval is going to be average rate of change of y with respect to x is going to be equal to, well, when x changed by 4, by positive 4, y changed by positive 2. So it's equal to 1/2. So it does look like the average rate of change over this interval right over here is 1/2. So we got lucky in this situation, our first choice-- and this is a multiple choice, so it's not a multiselect here. So our first one met our criteria. So we know that that's the answer. But let me do one more of these other ones to show you why that is not the answer. So let's find the average rate of change between that point and that point. So let's do another-- let me do it in another color. So I'll do this one in purple. So 0 is less than x, which is less than 4. And I'll just do the table right over here. x and y. So when x is 0, what is y? Well, it's going to be 1/8 times 0 minus 0, and y is just going to be 0. When x is 4, what is y? Well y is going to be-- let's see, it's going to be 1/8-- trying to do this in my head. 1/8 times 4 to the third. 4 to the third is 64. 1/8 of 64 is 8. It's going to be 8 minus 4 squared, which is 16. 8 minus 16 is negative 8. So in this one, x is increasing by 4, x has increased by 4, what happened to y? y has decreased by 8. So the average rate of change of y with respect to x here is y changed-- so I could write my change in y-- this Greek letter delta just literally is shorthand for change in. My change in y is negative 8 when my change in x is 4. So the average rate of change here is negative 2. It's negative because as x increased, your y decreased. And on average, for every 1 that x increased, y decreased by 2. That's where you get your negative 2 here. So this clearly, the average rate of change is not 1/2. So that confirms that that is not the answer. And we know that these other two, and I encourage you to try them out, will also give you an average rate of change of something other than positive 1/2.