Antiderivatives with Auntie Derivative

Hi everyone and I'll be your scribe for today.

We started today's class with some practice with the Mean Value Theorem. The following picture show the questions and the solution. I'll explain the work afterward, don't worry :)



Okay so for all Mean Value Questions similar to the ones seen in the homework and on the image above, you must check to see if the function is continuous and differentiable in the given interval. In other words, make sure the function doesn't jump values, go to infinity (vertical asymptote), or have any cusps or corners. One example of a function that may not satisfy the Mean Value Theorem would be:


This function is differentiable as explained during class, but one thing may prevent us from using the Mean Value Theorem on it. That one thing is the given interval in the question.



The red function is the graph of f(x)=|x|. The Mean Value Theorem says that the function must be continuous and differentiable within the interval. But as you can see the derivative of the function (blue) is not exist when x=0. So there for if the interval given in the question includes 0 in it, the absolute function does not satisfy the Mean Value Theorem.

Okay back to the questions. So let's look at the domain of the functions and the interval given in the function. The first question involves the natural logarithm function. From Grade 12 Pre-Calculus we know that the argument of any logarithm function cannot be less than or equal to zero because any base, except zero, raised to any exponent cannot equal 0 or negative value so since the argument of any logarithmic function is a power, the power cannot be zero or negative. So let's look at the permissable values of the argument.

As you can see the argument is a quadratic function. Now lets try to imagine the graph of that quadratic. Do you see it? Well let me help you out. Well lets forget about the constant (4) and complete the square with the remaining terms. You get:


Now let's imagine this graph for now. It is a parabola, opening up, but shifted to the left 1 unit. When the function is factored out we see that the quadratic in the argument of the ln has a constant of 4 so then that means that the parabola shifts above the x-axis. We know from Grade 11 Pre-Calculus that if a parabola opens upwards and has a vertex above the x-axis, it has no roots nor can it produce negative outputs. So we don't have to worry about the argument of the logarithm function from the question on the slide being negative.

Now that we've gotten that out of the way, the fun starts. So let's plug and chug the value's into the formula of the Mean Value Theorem and find where on the function does it's derivative match the Secant line connecting the two endpoints of the function in the interval. If you don't remember the formula of the Mean Value Theorem it is:


Once f'(c) is found we then make that equal to the derivative of the original function which can be found on the image but I'll make an image of it anyways (Because I like Sitmo):



So what you can do to solve for the variable is shown on the image with the question on it. You can, one, punch the derivative function into your calculator in Y1 and in Y2 insert your value f'(c) that you calculated earlier and find where the two lines intersect. Or you can bringing over f'(c) to the side with the derivative function as seen in the image with the question on it and graphing that, then find the roots. Now you have the answer.

For the next question, just follow the same process. Find out if the function satisfies the Mean Value Theorem within the given interval, apply the formula of the Mean Value Theorem, and find at what other input of the function will it have an output with the same derivative as the Secant line connecting the two endpoints of the interval. So the big thing to watch out for are the permissable values of the function.

Okay once that was over we began a new part of derivatives, the Antiderivative. Most of the time in our lives, when we learn to do something, we want to learn how to undo it. Let us take multiplication as an example. To undo multiplication you use division. In Mathematics that is called applying the inverse. If you remember from previous classes in Pre-Calculus, the inverse of any function undoes what the original function does. In Derivatives, when you want to undo "differentiating" the function, you "antidifferentiate" it. Let me bring out the image:



On this slide, we were asked to differentiate the functions above. All of their derivatives were the same, they were 2x. Okay cool, now lets go backwards, can you find the parent function when given it's derivative. Sure you can, you just apply the power rule in reverse. But there is one problem, how do you if there is a constant or not. It could be 1, 4, 34873, or pi to the power of 3i. When you differentiate the parent function, the constant disappears because of the constant rule of differentiation. The derivative of any constant is zero. So in general, we just add in c where c is a constant, as seen on the image. Okay lets look at the constant of any function, lets take x². When we add a constant to it, it causes the parabola to move up and down. All of those values may change the look of the parent function but they are all related in way because the derivatives are the same. So these functions are all related like a family.

We can check to see if our work was right by taking the derivative of the solution we made. If it's derivative was the same as the derivative given in the function, our answer is correct.

We were then asked to find the parent function if the derivative is x. Well the first guess was 1/x but when we applied the power rule to it, we didn't get the same derivative as in the question. Another guess was made and it was correct. (Good Job Jamie). So Mr. K gave began explaining the rule to finding the antiderivative, but the bell rang and class was over.

Okay let's recap. When working with the Mean Value Theorem, make sure there are not conflicts with the permissable values of the arguments of the functions and make sure the function is continuous and differentiable within the given interval.
We started the Antiderivative section of the Derivative Chapter and we'll continue it on Monday.

Oh Yeah, Remember Mr.K may do a sudden homework check when he feels like it. So get caught up with your homework or it's a zero for you. Just thought I'd remind you =D

Remember the test for Chapter 5 is on Wednesday, so that means the Pre-Test will be on Tuesday. We'll be starting the new Integrals unit on Thursday, and continue after the Holidays.

Well the next scribe will be my friend Jamie.

Good night and have a great weekend.