- Mean value theorem
- Conditions for MVT: graph
- Conditions for MVT: table
- Establishing differentiability for MVT
- Conditions for MVT: graph
- Justification with the mean value theorem
- Mean value theorem example: polynomial
- Mean value theorem example: square root function
- Using the mean value theorem
- Mean value theorem application
- Mean value theorem review
The mean value theorem applies to a function ƒ over an interval [𝘢,𝘣] under the conditions that ƒ is differentiable over (𝘢,𝘣) and continuous over [𝘢,𝘣]. See how we determine these conditions given a graph.
Want to join the conversation?
- At4:51, why is it differentiable over the open interval? And also, let a function which has an end point at c. Is the function continous at c?(2 votes)
- Why is the graph
hdifferentiable over the open interval
I believe differentiable pretty much means that if someone asked you "What is the slope of the line at spot?" you would know the answer, unlike at3:40.
So the function h is differentiable over the open interval
[-3, 2]because I can tell you without a doubt that the slope at any point in that interval is 1.
Let a function which has an end point at C. Is the function continuous at C? Continuous means there are no breaks in the function's graph, unlike in the video above at2:18.
The answer to your second question depends on what the graph looks like at C.
Hope this helps, and feel free to ask more questions!(1 vote)
- Does a function need to be continuous/differentiable to have one tangent line? Secant line?(1 vote)
- Find the number c that satisfies the mean value theorem for the function ((4/x)+x) on the interval [1,4](0 votes)
- 𝑓(𝑥) = 4 ∕ 𝑥 + 𝑥 is differentiable over the interval [1, 4], so the mean value theorem is applicable.
This means that there exists a 𝑐 ∈ [1, 4] for which 𝑓 '(𝑐) is equal to the slope of the straight line between the points (1, 𝑓(1)) and (4, 𝑓(4)).
First of all, the slope is (𝑓(4) − 𝑓(1)) ∕ (4 − 1) = (4 ∕ 4 + 4 − (4 ∕ 1 + 1)) ∕ 3 = 0.
Secondly, 𝑓 '(𝑥) = −4 ∕ 𝑥² + 1 ⇔ 𝑓 '(𝑐) = −4 ∕ 𝑐² + 1
So, we want to solve the equation −4 ∕ 𝑐² + 1 = 0 ⇔ 𝑐² = 4 ⇔ 𝑐 = ±2
However, 𝑐 ∈ [1, 4] ⇒ 𝑐 = 2(4 votes)
- [Instructor] So we're asked, does the mean value theorem apply to h over the interval? And they actually give us four different intervals here. So we should separately consider them. And this is the graph of y is equal to h of x. So pause this video and see, does the mean value theorem apply to h over any or all or some of these intervals? All right, now let's work through this together. In other videos, if the words mean value theorem are completely unfamiliar to you, I encourage you to look at the mean value theorem introduction, or the existence theorem introductions on Khan Academy. But as a review for those of you who are familiar with the mean value theorem, the conditions are, if we're dealing with some closed interval from a to b, we have to be differentiable, all right, just diff, actually, I'll just write it out, differentiable over the open interval from a to b, and we have to be continuous over the closed interval from a to b. Or another way to think about it, if you're differentiable, you're definitely going to be continuous. So, this second condition, just make sure that we're continuous at the endpoints of our closed interval. But if these conditions are met, then the mean value theorem tells us, and I'll just visually draw it because this is a review here. So let's say that this is our function. So let's say our function looks like this, this is a and this is b, and let's say that we meet these conditions over this interval, the mean value theorem tells us that there's going to be some value, let's call it c, in the interval, where the derivative of c is equal to the average rate of change from a to b. So the slope of the secant line is the average rate of change. And so you can see it visually, it looks like right about there, the slope of the tangent line would be the same. And so this would be the c that exists. Now, in this video, we're not trying to identify the c's, we're just trying to say, can we apply the mean value theorem? So let's look at this first interval from negative five to negative one, a closed interval. So we're going from negative five to negative one. So are we differentiable over the interval, over the open interval? We have this discontinuity over here that's not going to make it differentiable, and, of course, this is also, if we're not continuous, we're not going to be differentiable here. So this discontinuity here actually violates both of these conditions. And so for this first interval, we would say, no, the mean value theorem does not apply. Now, the second one, from negative one to three. So I'll do this in a slightly different color. So we're gonna go from negative one, right over there, to three. And if you look over this interval, it looks like, over that interval, over that closed interval, our function just really looks like a line. And so it is both continuous and differentiable over that interval, and it makes sense that the mean value theorem applies. Actually, every c on this interval is the derivative, is the instantaneous rate of change equal to the average rate of change because it looks linear over this interval. So the mean value theorem definitely applies over there. Now, what about from three to seven? The closed interval from three to seven. So when we look at this closed interval, we're definitely continuous over the interval. So we meet the second criteria, but are we differentiable? Well, a good giveaway of a point that is continuous but not differentiable is right over here. Whenever you have these sharp edges, because we don't have a well-defined slope of a tangent line here. Would it be that, would it be that, would it be that? And so because we aren't differentiable at that point, at x is equal to six, we aren't differentiable over this open interval from three to seven, and so the mean value theorem does not apply. And you can even see that visually. The average rate of change between the endpoints of our interval, or if we want to think about the slope of the secant line, that's that right over there, and over the interval, we don't see any point where the instantaneous rate of change, where the slope of the tangent line is equal to the slope of the secant line between the endpoints of our interval. So once again, mean value theorem would not apply. Now, what about from negative three to two? I'll do this in orange. So from negative three, right over there, to two. Well, if we look at the open interval from negative three to two, we're not considering what happens at negative three and at two, over the open interval, it does look like we are both differentiable and continuous. So we're definitely differentiable over the open interval. But clearly, we're not continuous over the closed interval. At negative three, we actually are not continuous. And so because of that, the mean value theorem does not apply. And you can actually even see that it would not apply because if you look at the slope of the secant line between our endpoints, that's the secant line right over there, and so you could see, over that interval, there is no c between negative three and two, where the slope of the tangent line or the instantaneous slope or the derivative is going to be the same as the slope of the secant line. And that doesn't violate the mean value theorem because the mean value theorem just doesn't apply here, we haven't met the condition of being continuous over the closed interval.