MEAN VALUE THEOREM (MVT) for derivatives (KristaKingMath)

In order to show that a function satisfies the mean value theorem on a particular interval, we'll use the mean value theorem formula. In order to use the formula, we have to know first that the function is both continuous and differentiable on the interval. If the function meets these two conditions, then we can apply the mean value theorem formula.

In order to find the value of c where the function has its mean value in the interval, we have to find the value of the function at the interval's endpoints. We'll also find the derivative of the function at the point c.

Once we have all of this, we'll plug everything into the mean value theorem formula, and that'll allow us to solve for the value of c that corresponds to the mean value of the function in the interval.

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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)

Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”

[rockwalldesmond]

You are a good math teacher. :)I can learn calculus by myself, of course I can't learn without your videos!

[urpaljp]

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. :)

[rockwalldesmond]

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!!

[mqk]

Nice video. Small correction though, the line that connects the end points is called a secant line.

[zuhairalsaffar]

Thanks you did your best l like your way of teaching all the best for you

[ewaldgroenewald]

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? :)

[XxxXxx-yhgz]

I wish my prof explains it as clear as you did. Thank you.

[elvismatemba]

Great explanation, flawless.Thank You Madam

[lyleoneal]

0:23 secant line containing the endpoints of the interval, not tangent line. 🙂

[nosaints]

Do you have any videos on differential equations? I'm in a differential equations class and power series is way over my head.

[jimkeller]

So what purpose does this serve beyond a mere mathematical exercise?

[fredflintstone]

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!

[nycandre]

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!

[husseinzedan]

i dont even need a school thank u
by the way i am arabic