This week we went over The first derivative test and we also learned how to do the second derivative test.
The second dervative test is basically the second derivate after the first derivative when we use the second derivative test we looked for: points of inflection, intervals of concavity, and we did the shortcut for max and min.
We also learned what differentiability is; differentiability is like a corner, a cusp or should I say BUTT, a vertical tangent, and anywhere it’s not continuous.
Well anyways an example of the second derivative test is:
6/x^2+3 the first derivative would be -12x/(x^2+3)^2 and to get the second derivative you would take the derivative of the first one and after deriving and simplifying the second derivative would be:
And the second derivative would be 36(x+1)(x-1)/(x^2+3)^3
But my question is can anyone help me find out the points of inflection, and the intervals and that stuff because I am kind of lost after that part. And Can some re-explain how to do the shortcut for Max and Min because I didn’t quite understand that.
But other than that I am pretty good but that which I just asked is most of the lesson so I’m really worried about the few test and quizzes we got coming up
Subscribe to:
Post Comments (Atom)
Ok, the shortcut for max and min. First you take the first derivative of the origional function and set it equal to zero just as you would do in the first derivative test. After you get your points, instead of making intervals out of them, you simply take the second derivative of the origional function and plug each of your critical points into your second derivative. After calculating this, you'd see if your graphs go +ve, -ve, +ve. If your calculations work out this way, then there are maximums and minimums at the points you pluged into your second derivative.
ReplyDeleteto find the points of inflection i'm pretty sure you have to take the second derivative, then set it equal to 0, then simplify or whatever, then you set up your intervals, then you plug in the numbers back into the second derivative, and if it's positive it's concave up, and if it's negative it's concave down. then you can find your points of inflection from there. for a more detailed description with an example go look at my post.
ReplyDeleteif you want to find the points of inflection, you have to take the second derivative. after that im pretty sure you set it equal to zero. then find interval and plug it back into the second derivative.
ReplyDelete