So this past week in calculus was a short one especially for me since I was only there 3 of 4 days. But the week started reviewing the first derivativetest. This has to do with the increasing, decreasing, positive slope, negative slope, concave up, concave down, horizontal tangent, slope=0, derivative above x axis, derivative below x axis, zero of derivative, maximum, minimum, and point of inflection in graphs. These things are used to get the derivative depending on which function they go with whether it being the original function, 1st derivative, or 2nd derivative. And the main thing these are used for is looking at a graph.
Then there are the steps to the first derivative test. The first thing to do is to take the derivative. Then set the derivative equal to 0. Next thing is to solve for x. After take the intervals from the last step and set them up. Then plug in the intervals into the first derivative. Finally in order to find an absolute maximum or minimum plug the values from the previous step into the original function.
I understand for the most part the first derivative test. But my problem is I went visit LSU Friday which means I missed calculus which means I missed everything we learned about second derivatives. Plus I just found out about a test on everything on Monday. But I would really appreciate if someone can explain the second derivative test to me before tomorrow so that I know it if it is on the test.
Subscribe to:
Post Comments (Atom)
The second derivative test is just like the first derivative test except you have to find the second derivative after you find the first and you use second derivative vocabulary instead of first derivative. so the steps would be
ReplyDelete1. first derivative
2. second derivative
3. set equal to zero and solve for x
the points you get are not critical values they are possible points of inflection
4. set up intervales and plug into the second derivate a number between. If its positive, it is concave up and negative is concave down.
5. Find points of inflection which is where the function changes concavity
you take the derivative of the derivate and then set equal to zero. Also, for your intervals it is the same thing too except positive would be concave up and negative would be concave down. That gives you your point of inflection and your maxs and mins.
ReplyDelete