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

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