Post #20

Goooooooooooodmorning! Since I just had to wish my sister to have fun on her vacation that she's leaving on today, there is nothing i would enjoy doing more but my homework being that we go to school tomorrow. YAY! (notice the sarcasm).

Anyways, i have another throwback blog because i didn't bring my binder home so i'm taking these lessons straight from the noggin. So here it goes.

The Second Derivative Test involves points of inflection and convavity, which is concave up or concave down. When taking the second derivative you must be aware that points of inflection only happen if there is a change in concavity.

To do the second derivative, first, you must take the derivative of the equation such as

6/(x^2 +3) = ( (x^2+3)(0) - [(6)(2x)] ) / (x^2 + 3)^2

Which gives you (-12x) / ( (x^2 +3) ^2) in it's simpliest terms.

Next, you must take the derivative of this equation.

Which would give you (36(x^2 -1)) / (x^2 +3)^3

This must then be solved to it's simpliest state set equal to zero which is
(36(x+1)(x-1)) / (x^2+3)^3

Now, you have found the critical values: X = +/- 1.

These are only POTENTIAL points of influction.
Next, you must set up intervals (-infinity, -1) u (-1, 1) u (1, infinity)

Now, you must pick a number in each interval and plug it into the second derivative. This will tell you if the interval concaves up or down.

f''(-2) = positive = concave up
f''(0) = negative = concave down
f''(2) = positvie = concave up

Another way to write this is:
at (-infinity, -1) and (1, infinity) = concave up
at (-1, 1) = concave down
points of inflection = x=-1x=1

These are the points of inflection because this is where the concavity changes. It is very important that you know points of inflection ONLY occur with changes of concavity.

I can't really remember any topics to say what i'm not understanding..so, hopefully everyone is having a good last day off, i might cry.

And umm, Juniors, did we have any homework?