Introduction to Anti-Derivatives (sounds like some sort of medicine)

So, on the previous class, we were introduced to anti-derivatives..


At the beginning of the class, we were asked to find the derivatives of the above functions. Needless to say, we found it difficult... NOT. It's 2x for all functions. And then we were given the derivative, f ' (x) = x and were asked to find the parent function. We were quite stumped at first, but then managed to figure it out. We had to go back to a previous lesson since we were asked whether if given a constant and a function and if we had to find the derivative of the function, does the constant play a role? No, it doesn't. Anyway, we were asked to find the parent function of the derivative f ' (x) = x. If x is the derivative, then obviously the parent function is x². But the derivative of x² is 2x and the derivative we want is x. So we figured, we have to multiply 2x by 1/2 to get x. From that, we got the parent function, which is 1/2x². Mr. K then asked us, what about if I give you the derivative x²? What would be its parent function? We then thought about it. If x² is the derivative, obviously the parent function is x³. But the derivative of that function will give us 3x², which we don't want since we're given the derivative x². So we thought multiplying 3x² by 1/3 would give us x². We reckon that the parent function is 1/3x³. From that we came up with a rule for all power functions, which is:

f ' (x) = xⁿ -------> f (x) = xⁿ⁺¹/n+1 + C, where C is the constant. I'll explain later.


We then made anti-derivative rules for almost all derivative rules we remember. Mr. K said that the anti-derivative rules for the product and quotient rule are much more complicated and we won't be taking it until First Year University Calculus, so yeah.


Now this is where I explain the C part in the rule:

f ' (x) = xⁿ -------> f (x) = xⁿ⁺¹/n+1 + C, where C is the constant

We can't really find C not unless we're given a point in the graph. C is basically the y-intercept, but if we're just given the derivative and asked to solve for the parent function, all we can do is find the parent function and then + C, not unless we're given a point in the graph.

Then we were just given a practice problem, which was pretty straightforward.






The next scribe is etimz, or Ethan.

Today's Slides: May 23

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BOB

Well, another test. This unit hasn't been so bad for me, I understand most of it. Hopefully I do well in the test. I do have a couple of problems with differentiating implicitly but it's nothing practice cannot solve. Otherwise, I'm good to go.

Good luck.

BOBBOBBOB!

What can I say? Well for one, this unit was simple the second time around =). I really think that I'm going to do well on this next test and yeah, hopefully I don't stumble on something that I haven't yet seen or tried to do. As always, the solution is practice! Hmm, I think the best place to start review would be to go through the past slides, and then see if I have any muddiest points.

This bob was short and straight to the point. I hope everyone remembers to bob and and and! Good luck on the test Wednesday I believe!

Today's Slides: May 16

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Putting the pieces together

Today's lesson was about putting the pieces together, so to speak. Mr. K taught us how we should be able to come up with a graph of a function simply by using derivatives and critical points and whatnot.


This question was pretty straightforward. It just asks us to determine the critical points of the function, which means we have to find out where the derivative = 0. First we have to find the derivative of the certain function, as shown above, factor it if possible. However, don't make the mistake I did, which is state the CRITICAL NUMBERS instead of stating the CRITICAL POINTS. To find the critical points, just plug in the critical numbers at the parent function, and you should have it.



This is another straightforward question. Mr. K said that we should be able to solve these kinds of questions for 30-60 seconds. To find the relative max and min of an equation, we simply have to find out where the derivative of the parent function = 0. We then find the derivative of the parent function, as seen above, and then we use the first derivative test to find out if we have a maximum or a minimum. Then, we solve for the second derivative and use the second derivative test to find out again the maximum or a minimum of a function. An interesting tidbit: the second derivative function's main purpose is to find out the maximum or a minimum of a parent function. If we're only given the second derivative of a function, then yes, the 2nd derivative function is useful. Then Mr. K said, but what if we're given the parent function? Isn't that kind of a waste of time? The first derivative test already tells us the maximum or a minimum of a function, so why bother with the second derivative test? If it's main purpose is to find out the concavity of a parent function, then there we go.


For this question, we had to find out several things; intervals in which y is increasing or decreasing, the coordinates of any local extrema (minimum and maximum), intervals of concavity, inflection points, and the graph of the function. To find out where y is increasing or decreasing, we have to find out its derivative first and test the answers that we get. As you can see on the picture above, the critical numbers are x = 1 and 2. f is increasing where f ' is positive, and f is decreasing where f ' is negative. We then test it by using the number line test, and the

To find the coordinates of the local extrema, we use the first derivative test. From the number line test we did, we can see that we have a max at x = 1 and a min at x = 2. We then plug in those critical numbers on the parent function to find out the y coordinate of the maximum and the minimum points.

To find the intervals of concavity, we simply use the second derivative test. We find the derivative of the first derivative, which of course gives us the second derivative. We then find its critical point. We then use the number line test to find out where the parent function is concave up and concave down. From the solution above, we find out that the parent function is concave down from (-∞, 3/2) and concave up at (3/2, ∞). Therefore, we can say that we have a point of inflection at x = 3/2.


Well this is pretty straightforward. We just gather what we found out from the previous solutions and, shall I say, "put the pieces together." Then we sketch the graph. =)

That's it! We have a pre-test next class and a test the class after.

The next scribe is... *drum roll please*

haven't decided yet. lol

Today's Slides: May 14

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