hmm, so this is the first post of the new year, but it does not feel any different from all the other posts. oh well... this post is going to be on the second derivative test. The second derivative test is used to find all possible points of inflection and intervals 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
i knew how to do pretty much everything for the exam, so the only thing i would need to know is maybe optimization and just a review of some other stuff
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Optimization... You are usually working with two equations, a primary and a secondary. The primary will be the equation of whatever you are optimizing...so for example, find the maximum area... would use the area formula as the primary equation. As for the secondary...it's simply the other one.
ReplyDeleteSteps are...
Solve the secondary for a variable. Plug this into the primary so that it is now in terms of one equation. Now, take the derivative and set equal to 0 because you want to find the max or min of it. After solving, you can plug in the values, including plugging in your bounds and check to see which one is the maximum or minimum. This will be your answer.
Feel free to see me in class or after class for more help on this :-)
Optimization is pretty easy when everything's given to you in the problem. In the problem, you will be given two equations and you are asked to maximize or minimuze one of them. The one you are asked to maximize or minimize will be your primary equasion. Your secondary will most likely be set equal to a number. These are the steps:
ReplyDelete1. Locate the primary and secondary equasions
2. Solve the secondary for one variable
3. Plug into primary
4. Take the derivative and solve (yielding one of your answers)
5. Plug back into your secondary equation to get your second answer
*Usually in optimization equations they are asking for more than one answer. For example, if they are asking to maximize an area, they will most likely ask to find the perimiter of an object, yielding more than one answer.
So, I would try to help explain optimization, but I'd feel as if I'm stealing the work of others. It's explained four different ways on my comments, so you might want to refer to those :D It helped me LOTS!!
ReplyDeletehttp://br0910apcalc.blogspot.com/2010/01/ashs-20th-post.html#comments
As they said you are working with two equations, usually area or volume or an equation like those. The secondary equation is one that you will usually derive from what the problem is asking for if its not given.
ReplyDeletefirst: set up primary and secondary equations
Second: solve secondary for a variable.
Third: plug into primary and take derivative.
Fourth: solve for variable.
Fifth: plug back into the secondary equation and solve for other variable