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That is why I ended it with pre-cal. *points to le first post*

Man, I'm in the middle of taking second derivatives right now. My class has to take a function and graph it without a calculator and within enough time that it could be conceivably done on the AP Exam (i.e., without testing eight zillion points). We have to...

1 ) Find zeros
2 ) Find y-intercept
(Not bad so far)
3 ) Find all asymtopes
4 ) Find derivative
5 ) Find all critical points
6 ) Check for relative max/mins
7 ) Draw sign chart for derivative
(Still, really not much of a problem, though asymtopes are a pain sometimes)
8 ) Find second derivative
9 ) Find all possible points of inflection
10 ) Draw sign chart for second derivative
11 ) Find all real points of inflection
12 ) Draw graph
(This is where it gets sketchy *haha* because the second derivative of a fractional or trigonometric function is nasty)

How many of you sympathize on this point? I'm not trying to be arrogant, if that what it looks like; I'm trying to figure out where people stand in whatever course they're taking. Just to give an example...

f(x) = (2x - 3)/(x^2+1)

And we have to do three like that in a fifty-minute class period. It takes for-freaking-ever!

We've been doing that for a while now. *Test 2 coming on Thursday*

Lol, that example above is nothing compared to trig. At least you know, approximately, what it's going to look like, whereas for trig, we have to memorize every single function.... and to that our prof had to add hyperbolic functions as well.

Ayeeee.... someone's going to fail that test on Thursday crying

im in grade 10 and iv dun this but i amin the gifted class.. but not for my spelling sweatdrop

No problem! It's always good to know the basics before moving onto more difficult stuff - I was totally clueless in grade 11 AP physics because my grade 10 teacher decided to just give us a bunch of formulae to memorize, and not teach at all. So yes, always make sure you get the basics, especially since it'll only get harder and harder. Best of luck to you =)

Woot, I can finally take my calulus knowledge and use it to help others 3nodding


Extrema are simply the high and low points on a graph (ex: the min of y = x^2 is (0,0); there is no max however since it goes on to infinity).

Finding extrema is easy say you have a function f(x), you'd take the derivative of this function and solve for zero. The reasoning behind this is simple, a derivative is simply the slope of a given equation (the slope of f(x) = f'(x)). Since a maximum or minimum point on a graph has a slop of zero, you'd need to find the zeros for the equation f'(x).

Now that you have the points where f'(x) = 0 (aka, the critical points), you need to test them out to see which is a max or min. This is also fairly simple, you make a simple line graph with the critical points on them. Now you take values that are higher or lower than these points and find out if they're positive or negative. If the pattern goes POSxNEG then the point is a max, if it goes NEGxPOS then its a min. To get the coordinate, just plug the critical point (this is your x) into the original equation to find your y.

Now for an example!
f(x) = x^2 - 5
f'(x) = 2x
2x = 0
x = 0, Critical Points

<-------0-------> (poorly made line graph sweatdrop )
Testing, x= -1 and x= 1
f'(-1) = -2, NEG
f'(1) = 2, POS

<--0++>
Since to the left of 0 the graph is negative, and the right positive, this makes the point x= 0 a minimum. Now plugging x back into the original equation gives:

f(0) = (0)^2 - 5
f(0) = -5

Giving the coordinates of our minumum (0,-5).

This same procedure can be used on any equation, this is easy stuff... just wait until you get to integrals,

that's weird... i'm in grade 10 and i've already learned max and mins

oh c'mon kid we are not here to do your homework!