second reflection

okay, so for my second reflection i am going to go into detail on the second derivative test to find all possible points of inflection and intervals of concavity. remember, points of inflection only happen where there is a change of concavity.

Example: f'(x)= 6/(x^(2)+3)

First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative.

Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative.

Once again u need to use the quotient rule, so f''(x)={(x2+3)^2-(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, so I am just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^2+3)^3

The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1

so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity)

then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value

then you know that your intervals concave up at (-infinity, -1) u (1,infinity) or x<1,>1

and it is concave down at (-1,1) or -1
and you're points of inflection are x=-1, and x=1

so that is all for the second derivative test, it's not that hard, but some problems have a bunch of quotient and product rule and other stuff, so they can take forever to do, but hopefully we don't get any of those on the exam.