This week we took ap tests yet again, and corrected them. so i am going to do my blog on the first derivative test. the first derivative test is used to find absolute maximums and minimums, to do these, first yhou take the derivative of the equation given to you, then you set it equal to zero and solve for x, then you set up intervals using the x values, then you plug in numbers within the intervals into the derivative to find where your relative maximums and minimums are, then you plug the x values into the original equation to find the absolute maximums and minimums.
So here is an example problam i had:
Find the absolute maximums and minimums of the function f(x)=x^2-8x+4
First, you take the derivative of it:
f'(x)= 2x-8
Then you set it equal to zero and solve for x.
2x-8=0 2x=8 x=4
Then you set up the intervals
(-infinity, 4) U (4, infinity)
then you plug in a number found within the interval into the first derivative to find relative max's and min's.
(-infinity, 4)= negative number
(4, infinity)= positive number
so at x=4 there is a minimum
after that, you plug in that x value into the original equation to find the absolute max's and min's. (in this case it is only a min)
x=4 is the absolute minimum (4,-12)
i could use some help on substitution..
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Substitution:
ReplyDelete1. State your u and du
2. Integrate
3. Plug u back in
also a lot of the time, you need to substitue in a number. when doing this, simply put it in front of the problem when you are done.
The only thing I can add to Jessie's explanation is that in order to find u and du, look for something that's being integrated or close to it (as Jessie said, you can add numbers if it doesn't work) and that'll be your u...I hope that didn't make it worse!
ReplyDeletesubstitution:
ReplyDeletefind your u & du.
then integrate.
plug u back in
it sounds confusing, but it's easier than it looks.
u is usually whatever is in the paranthesis.
du is the derivative of u.
also, du eventually cancels out whenever you integrate.. that's why you only plug u back in.