Revision of Mean Value Theorem

Well, I'm too lazy to go back and change my first post, so I'd rather just make a new post, which contains only the revisions, not the entire post all over again. If anyone wants to see the first post for some reason, the link should be on the right side of the screen somewhere.

Why not-differentiable and discontinuous functions fail the Mean Value Theorem:

Ok, so apparently why they fail needed some revision, so let's try this again. Let us think of the function f(x)=x^2, because everyone knows how it looks. We all know it passes the Mean Value Theorem, but let's see why. There are no discontinuities, cusps, corners, or anything of the sort. As we get closer to x=0, the derivative of f gets closer to 0, and as we leave 0 towards the positive side, the derivative becomes positive. That is the general reason why differentiable functions are guarenteed a point where slope of the tangent line at x=c equals the slope of the secant line: there is a smooth transition in the sign of the derivative. If we are talking about straight line, the tangent line is the line itself, so there is no worry. Ugh...explaining with words is becoming difficult, so lets look at pictures.



Ok, again, I'm terrible with the size of picture, so I'll just explain what is happening. The graph is f(x)=x^(4/5) and the interval we are considering is [-1,1]. The secant line is y=1. There is a cusp at x=0. Because it's continuous but not differentiable, it fails the Mean Value Theorem. Why? If we examine the derivative of f, we see that it's negative at x<0>0. But if we think back to f(x)=x^2, it lacks a vital charecteristic. There is no smooth transition in the derivative from negative to zero to positive, as all curves have. In this graph, the derivative is negative, goes to a point without a derivative, and then is suddenly positive. The smooth transition just isn't there. That is why there is no guarenteed value for which the slope of the tangent line equals the slope of the secant line. I'm guessing the explanation is still a little fuzzy, so let's look at another picture.



This graph is the graph of a piecewise function, composed of y=x^2+1 from [-1,0] and y=x from (0,3], where there is a jump discontinuity at x=0 and has a secant line of y=x/4+(9/4). We know it fails the mean value theorem, but why? The exact same reason as the example above. We don't see that smooth transition in the derivative from negative to zero to positive. It goes from negative to a point where there is no derivative (due to the discontinuity), and then the derivative is thrust into being positive. The derivative doesn't naturally change but changes abruptly and suddenly. It misses values for the derivative, values the derivative can't be without the smooth transition (same as in the example above) which is why functions that aren't differentiable or continuous can't have a guarenteed point where the slope of the tangent line equals the slope of the secant line.

Wow, that was a lot of writing. That should explain everything.
(P.S. You should put more videos like that, Ms. Hwang. That contraption was extremely entertaining.)

Mean Value Theorem

Wow, so much more work for a blog post. Not used to that...

1. To understand what the mean value theorem means, you need to be able to translate it into English. f '(c) refers to the derivative at x=c and [f(b)-f(a)]/(b-a) refers to the slope of the secant line through the interval [a,b]. The mean value theorem, therefore, means that if there is a differentiable function on the interval [a,b], which implies continuity, then there is a guarenteed point at which the slope of the tangent line equals the slope of the secant line for that interval. For my example, I used the equation f(x)=4cos x.

If we consider the interval to be [0,4pi], then there are numerous points in which the slope of the tangent line equals the slope of the tangent line. The slope of the secant line is 0, and so all max and min in the interval have the same slope of 0 (the calculations are easy enough, so I leave them all to you).

2. I'm not 100% sure why it only works for differentiable functions, but my reasoning is this. If the function is both continuous and differentiable, then you can get any number as a derivative, either large or small (assuming the function is a curve). If the function is a line, then there's only one possible derivative, so by all points would have the same slope as the secant line because the function is the secant line. If the function is either not differentiable (cusp, corner, discontinuity, vertical tangent) or discontinuous, it misses that crucial charecteristic that curves have, it being that the derivative can be wither large or small. Yes, my explanation isn't the best, so lets look at a case of each and see why they fail.


This is continuous but not differentiable: f(x)=abs(x). If we pretend the interval is [-5,5], then the slope of the secant line is 0. However, there are only two possibilities for derivatives in the actual graph: -1 and 1. Relating back to what I said before, it doesn't have the charecteristic that curves have that the derivative can be large or small.



The graph is f(x)=abs(x)/x (yes, I know the graph looks strange, but it's the calculator). If we take the interval to be [-1,1], on which the function isn't continuous, then the slope of the secant line is 1. However, the only option for the derivative is 0. Again, the graph lacks the charecteristic of all curves, in that the derivative can be anything, large or small.

(Note: Sorry if the pictures aren't clear, I somewhat experimented on their size.)