See Math

Continuous, but not differentiable.


This is a function that is not one to one.


Not continuous and not differentiable function.


Invertible Functions


A continuous and differentiable function.

A Very Late Choice Indeed

Derivative of an Inverse Function

The property is simply the reciprocal of the derivative of the inverse of the original function. In layperson terms, it is simply one divided by the derivative of the inverse function. Here it is in symbols to hopefully make it clearer.


This is the inverse of the volume function:

Below is the picture and description of the following functions.
(1) The inverse volume function r(V);
(2) The volume function, V(r);
(3) x=1 on V(r) and graph the tangent line through at this point.
(4) The tangent line through the “mirror” point on the graph of the inverse volume function.

The first function is the inverse volume function which is the gray function in the image of the graph. The second function is the original volume function which is the red function in the image of the graph. The third function is the tangent line to the volume equation at the point x=1. The forth function is the tangent line to the inverse volume function at the mirror point.

To find the function of the tangent line to point x=1 for the volume function you can simply type in the volume function into your calculator and use nDeriv(y1, x, x) to find the derivative at that point. Use the point slope form of a linear equation to form the tangent line at the point x=1 for the volume function. Use the derivative at the point x=1 for the volume function as the slope of the tangent line. Then insert the values of the volume function for x and y when x=1 into the point slope form as x and y. Aha, now you have the slope of the tangent line at the point x=1 for the volume function!

The function of the tangent line of the mirror point for the inverse volume function is found similarily to the previous tangent line. Because we know that this function is the inverse to the volume function the slope of the tangent line through the mirror point will be the inverse of the slope of the linear function of the tangent line to the volume function when x=1. Also we know that the x and y value at the point x=1 will be switched around for the tangent line to the inverse volume function at the mirror point. This is all that you have to do to find the function of the tangent line at the mirror point while utalizing point slope form.

I choose not to go to class and to take the virtual class because I wanted to try something new. Honestly I am quite biased and believe that you learn much better in the classroom instead of on your own by using the internet. I wanted to see if the classroom actually did have a big advantage.

Differentiability and Continuity

Hey guys and gals! Hope your all enjoying the new snow. Here is the scribe post for our class on Jan. 26, 2010.

Hope fully by the end of this post you will be feeling like one of those :)

To start of the post with an absolute value attitude here is a math joke for all of you fella's who like those nerdy pick up lines.

"I don't know if you're in my range, but I'd sure like to take you home to my domain."

Now back to business...

Please take a look at the slides from class today, they are important for understanding the post. It has pictures of the worksheet we did in class, which is debreifed in this post. If you loose your homework, you can either get organized or just look at the slides from today.

Today in class we discussed differentiability and continuity. We practiced our understanding of these two topics by creating graphs under certain guidelines. The following is a list of the guidelines we had to follow for creating our first 5 graphs:

Draw a graph that is...

1. Differentiable and continuous at point c.
2. Has a limit as x approaches c, but fails to be continuous there due to f(c) being undefined.
3. Has a limit as x approaches c, and has a value for f(c), but fails to be continuous there
4. Has a value for f(c), but has no limit as x approaches c
5. Is continuous at the point (c,y), but is not differentiable at this point.

I added these problems from our worksheet for anybody who wants to be an ogre achiever (ha, get it, ogre) and continue to use them as practice for the chapter 4 test. Seeing as I have ESP I predict that test is looming very near, sorry guys. I think these are a great way to do something similar to gymnastics for your brain (as Sara used to say for those of you who were in her class last year).

So whats the whole point of this?
The whole point of this is to come up with some rules that dictate when a function is non-differentiable at a point x=c. Here are the rules I came up with.

1) If there is no value for the function at point x=c, it is not differentiable there
2) If there is no value for the derivative function when f'(c), then it is not differentiable at that point
3) If f'(c) is not a real number, then it is not differentiable at that point.
4) Examples of non-differentiable characteristics that can be found at c: cusp, asymptote, hole, discontinuity.

If all you people of Calculopolis out there have any additions or modification to this leave a comment and I will edit this post accordingly.

Another important topic to discuss from class today was problem number 6. Problem number 6 gave us two functions, the first function

was to be within the domain
and the second function (listed below)
was to be within the domain of
Yet this piecewise function also had continuous and differentiable. Our Quest was to find the values of a and b

Sorry for the visible formatting issues here.

The first step to find the value of these two variables is to derive the first function (x^2). The derivative of the first function is 2x. When we substitute in the last number in the domain of the first function's derivative we end up with 2(6) which equals 12. Next, substitute 12 in for the variable a. This is to ensure that the slope of the linear function matches up with the slope of the first function, x^2 at the point x=6. This is to prevent a hole and keep the instantaneous rate of change (the derivative) the same when x=6 for both functions. Because b controls the vertical translation of the linear function, we have to make it match up with the original function x^2. To find where the end of x^2 is, you substitute in 6, which gives you 36. We now know how high the linear portion of the function has to be in order to match up with the first function x^2. The variable b we now know equals 36. Ta Da! We have now successfully solved the problem.
This essentially conclueds our lessons of the day.

Hopefully the world of Calculoplis is a little easier to understand now.

Thanks,
mc Casper

p.s. almost forgot, I dub J-tron foteen as our next scribe

21st Century Education

In the 21st century of course will be filled with unexpected new technologies that will demand new skills that we have never fathomed. Despite the incomprehensible digital expansion that we are witnessing, it does not completely and irrevocably alter the way in which we ought to teach. Personally, I think that education is much more than just learning material, but an experience. I believe that there is irreplaceable value in being a class room with a structured school schedule. The value of physically being in class could never be replaced by sitting behind a computer screen.

With all of the global issues going on today the most important skill that you will ever learn is problem solving and relating and communicating with people. The class room could be said to somewhat represent the workplace, where there is a manager who resides over people and tries to ensure their quality of work is of the best can possibly do. Similarly, a teacher is there to ensure the highest quality of learning for the students that they can provide and hence ask the students for their full participation.

When you are sitting behind a computer you do are unable to directly communicate with someone face to face and therefore lack the means to develop the communicative people skills necessary in most all aspects of life and the workplace. Also, I honestly think that the relationship between teacher and student when would be virtually eliminated with a completely online education. All in all, there are pro's and con's to both. I would hazard a suggestion at a mixture of the two types of education, with a foundation in class room type educational settings generally.

Product Rule

Personally, I found the website to be much better source for learning the product rule quickly and efficiently without compromising understanding. The book had much more lengthy and complete explanation of the product rule including graphs, verbal definitions, and symbolic definitions. Despite the thoroughness of the text book definition I still found the explanation on the website (http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html) to be much more useful for a person like myself. The online resource gave one short definition, but many full length examples of increasing complexity. I found this to be more useful than the textbook for mainly that very reason. It is often easier for me to understand a new concept by looking at multiple examples that I can easily refer to if I find myself stuck on a problem. Another additional reason as to why I prefer the online resource is because it is simpler and makes the product rule in calculus look less daunting while maintaining the information necessary. Hence, the online resource, oddly enough, gives me somewhat more confidence in my understanding of the concept.

Product Rule:
The product rule is a formal rule for differentiating problems where one function is multiplied by another. The rule follows from the limit definition of derivative and is given by where D stand for the derivative of a function (excerpted from http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html)

Example: Differentiate

Concepts of Calculus Mind Map