Thanks for doing this video and the video about the marginal, revenue, and profit videos. I took Calc I about a year ago and those two topics were the ones that confused me a lot. I'm heading into Calc III this semester and this channel is great for reviewing. :)
HonestLiar Yes, I have some DE videos. Check out videos 144-170 in this playlist, or just search my channel. czcams.com/play/PLJ8OrXpbC-BMdeuQfJDVRJ5DPMduSzVow.html
Can you do a video like this (MVT) on multivariate functions? I'm stuck at trying to convert my multivariate function to a single variable function in order to apply the MVT. Help!
Ask Google, the MVT doesn't generalize easily to higher dimensions. There are various extensions proposed, see for example www.math.caltech.edu/~dinakar/08-Ma1cAnalytical-Notes-chap.2.pdf etc
Sorry but this is not helpful at all - It is actually going to confuse people who are learning math. (1) You name this video "MEAN VALUE THEOREM FOR DERIVATIVES" and (2) you state as question "Show that the function f.. satisfies the MVT on ..[..]. For (1) the video has nothing to say about the MVT for derivatives, only for a specific function, and certainly not its derivative. So this is an incorrect title. For (2), more importantly, the statement on the blackboard is problematic, or at best confusing. What you really wanted to say is MVP ie Mean Value Property, instead of MVT. By MVP is meant the statement of a function satisfying "there exists c in the closed interval [a, b] such that f(b)- f(a) = ... WITHOUT any conditions on f being continous and differentiable. Or, to put it differently, a more appropriate title should be "ILLUSTRATE the MVT in the case of the function f(x)= 3x^2 + .. by calculating the value of c for the interval mentioned" The answer to the mathematical question as you phrase it would be: "This function satisfies the MVT because it is continuous and differentiable on any closed interval". PERIOD. If you were nitpicking you could add, also, that is because products and additions of continuous differentiable functions are also continuous and differentiable on the same intervals. The whole point of the MVT is to state that given conditions ( ..1... ) there ALWAYS exists a number c in the interval (..) for which (....). So for any function satisfying the given conditions ( ..1.. ) there WILL be a point c etc etc. To emphasize, the essence of mathematics is to be PRECISE, especially in the formulation of its discussions, proofs and questions. I took the trouble of writing this for a friend of my niece who thought this was clear. It is clear AND misleading. Please amend this! PS. Explanations and examples that are guaranteed not to be misleading can be found at Khan Academy or Wikipedia. In this case see for a much better explanation www.khanacademy.org/math/ap-calculus-ab/derivative-applications-ab/mean-value-theorem-ab/v/finding-where-the-derivative-is-equal-to-the-average-change
Hi Andre, Thanks for the feedback, I appreciate it. I'm sorry the video wasn't what you were hoping to see, and that you felt like it was misleading. My goal with the video was not to give a technical proof of the MVT, but really just to show people how to solve these kinds of problems specifically. As you know, this exact question "Show that the function satisfies the MVT..." appears all the time in textbooks, and I'm trying to show to steps you can take to solve this specific kind of problem, plus give a little background. Thanks again for sharing this extra information so that other people can dive in deeper if they want to.
Thank you!!!!!!!!!!!!!!!!! Cannot appreciate this more
Thank you, Krista!
You are a good math teacher. :)I can learn calculus by myself ,of course I can't learn without your videos!
姜炜veen 沈joh Aw thanks, I'm so glad the videos are helping!
I wish my prof explains it as clear as you did. Thank you.
Thanks for doing this video and the video about the marginal, revenue, and profit videos. I took Calc I about a year ago and those two topics were the ones that confused me a lot. I'm heading into Calc III this semester and this channel is great for reviewing. :)
Pbnjp You're welcome, I'm so glad the videos are helping! Best of luck with Calc III, I hope it goes great!!
Well done, flawlessly.
Marion Deleon Thank you very much!
Thanks you did your best l like your way of teaching all the best for you
Thank you so much, Zuhair!
I hope I can learn more from your amazing videos,I also hope you could have more videos for us. I will keep watching and learning!!
Great explanation,flawless.Thank You Madam
Thanks, Elvis, I'm so glad you liked it! :)
0:23 secant line containing the endpoints of the interval, not tangent line. 🙂
Nice video. Small correction though, the line that connects the end points is called a secant line.
Thanks, this help so much! We're doing the same theorem in the integral form. If possible can you make a video, explaining this form of the MVT? :)
you are a god.
Love u💙💙💙 really appreciate it
So what purpose does this serve beyond a mere mathematical exercise?
Do you have any videos on differential equations? I'm in a differential equations class and power series is way over my head.
HonestLiar Yes, I have some DE videos. Check out videos 144-170 in this playlist, or just search my channel. czcams.com/play/PLJ8OrXpbC-BMdeuQfJDVRJ5DPMduSzVow.html
Can you do a video like this (MVT) on multivariate functions? I'm stuck at trying to convert my multivariate function to a single variable function in order to apply the MVT. Help!
Ask Google, the MVT doesn't generalize easily to higher dimensions. There are various extensions proposed, see for example www.math.caltech.edu/~dinakar/08-Ma1cAnalytical-Notes-chap.2.pdf etc
*applause
Sorry but this is not helpful at all - It is actually going to confuse people who are learning math. (1) You name this video "MEAN VALUE THEOREM FOR DERIVATIVES" and (2) you state as question "Show that the function f.. satisfies the MVT on ..[..].
For (1) the video has nothing to say about the MVT for derivatives, only for a specific function, and certainly not its derivative. So this is an incorrect title.
For (2), more importantly, the statement on the blackboard is problematic, or at best confusing. What you really wanted to say is MVP ie Mean Value Property, instead of MVT. By MVP is meant the statement of a function satisfying "there exists c in the closed interval [a, b] such that f(b)- f(a) = ... WITHOUT any conditions on f being continous and differentiable.
Or, to put it differently, a more appropriate title should be "ILLUSTRATE the MVT in the case of the function f(x)= 3x^2 + .. by calculating the value of c for the interval mentioned"
The answer to the mathematical question as you phrase it would be: "This function satisfies the MVT because it is continuous and differentiable on any closed interval". PERIOD. If you were nitpicking you could add, also, that is because products and additions of continuous differentiable functions are also continuous and differentiable on the same intervals.
The whole point of the MVT is to state that given conditions ( ..1... ) there ALWAYS exists a number c in the interval (..) for which (....). So for any function satisfying the given conditions ( ..1.. ) there WILL be a point c etc etc.
To emphasize, the essence of mathematics is to be PRECISE, especially in the formulation of its discussions, proofs and questions.
I took the trouble of writing this for a friend of my niece who thought this was clear. It is clear AND misleading. Please amend this!
PS. Explanations and examples that are guaranteed not to be misleading can be found at Khan Academy or Wikipedia. In this case see for a much better explanation www.khanacademy.org/math/ap-calculus-ab/derivative-applications-ab/mean-value-theorem-ab/v/finding-where-the-derivative-is-equal-to-the-average-change
Hi Andre, Thanks for the feedback, I appreciate it. I'm sorry the video wasn't what you were hoping to see, and that you felt like it was misleading. My goal with the video was not to give a technical proof of the MVT, but really just to show people how to solve these kinds of problems specifically. As you know, this exact question "Show that the function satisfies the MVT..." appears all the time in textbooks, and I'm trying to show to steps you can take to solve this specific kind of problem, plus give a little background. Thanks again for sharing this extra information so that other people can dive in deeper if they want to.
i dont even need a school thank u
by the way i am arabic
I'm African 😂