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, and if I typed all that up it would be ridiculous, so I'm 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 @ (-infinity, -1) u (1,infinity) or x<1,>1
and it is concave down @ (-1,1) or -1
and you're points of inflection are x=-1, and x=1
okay, and now for my question, In the example mentioned above, I understand what you do to get the points of (-infinity,-1) u (-1,1) etc.... but then it says to have to take the second derivative of (-2), (0), and (2), and you do this to find out if the value negative or positive, but where do those numbers come from? The -2, 0, and 2. This is probably a dumb question, but thanks to whoever answers it.
you have to pick a number between the
ReplyDelete(-infinit,-1) which the easiest to use is -2
and its the same for the other two intervals
a number between (-1,1) is 0
and a number between (1,infinity) is 2
The -2, 0, and 2 are there because you plug back into the equation in order to see what intervals do what. You pick -2 because your first interval is (-infinity,-1) and that is the easiest number to pick. The next interval is (-1,1) which is why you pick 0, then the last interval is (1,infinity) therefore you pick 2. After plugging back into the second derivative you find where the graph is concave up and concave down based on whether it is positive or negative.
ReplyDeletewe picked -2, 0, and 2 because that's the
ReplyDeletenumbers we chose on when we were working
it together, but don't think that's the
only numbers you havce to use.
for the first interval you can choose
any number bbetween -infinity and -1.
the same thing goes for the third interval,
but for the second interval you must choose
zero because that is the only point between
-1 and 1.
plugs whichever numbers you choose into the second derivative to see whether it gives you
a negative or positive answer, you really
don't even have to do the math, just
look at the signs.
for that problem in class you coulda used any number. the only thing we were lookin for was that the slope was decreasing and there was only graph with a negative horizontal line was the answer
ReplyDelete