Welcome Guest [Log In] [Register]
    Welcome to Nintendo Forums. We hope you enjoy your visit.


    You're currently viewing our forum as a guest. This means you are limited to certain areas of the board and there are some features you can't use. If you join our community, you'll be able to access member-only sections, and use many member-only features such as customizing your profile, sending personal messages, and voting in polls. Registration is simple, fast, and completely free.


    Join our community!


    If you're already a member please log in to your account to access all of our features:

    Username:   Password:
    Add Reply
    • Pages:
    • 1
    • 2
    I would suggest a clock, and a support team.
    Topic Started: Apr 23 2005, 03:35 PM (495 Views)
    Ganondorf718
    Bombette
    Would you like some calculus and latin lessons for work? :)

    Hers a quick lesson:

    Derivative of a Polynomial

    Taking a Derivative:

    A derivative is basically a formula for finding the slope of a curve. When you take a function’s derivative, you are finding another function that provides the slope of the first function. Finding a derivative is simple for some functions, but it can get more complicated as we move on.

    First, we will take the derivative of a simple polynomial: 4x^2 + 6x
    The first step is to take any exponent and bring it down, multiplying it times the coefficient. In other words, bring the 2 down from the top and multiply it by the 4. Then reduce the exponent by 1. The final derivative of that term is (4*2)x^1, or 8X.

    The second term is 6x. Since the exponent is assumed to be 1, we can bring that down and multiply, which does not change the coefficient. Reducing the exponent by 1 makes it 0, so the derivative of 6x is just 6x^0, or the number 6. For any linear term (a number times a variable to the first power) like 6x, the derivative will simply be that coefficient.

    Remember, the original polynomial we were “differentiating” (taking the derivative of) was 4x^2 + 6x. Going term by term, the derivative was determined to be 8X + 6. That’s it!

    Now let’s take the derivative of a few more polynomials to make sure we understand the basics:

    x^2 + 8X + 13
    Derivative = 2x + 8 (note that any constant is eliminated, because essentially that was 13x^0, and when the 0 comes down the whole term becomes 0)

    3x^2 + x + 9
    Derivative = 6x + 1

    4x^4 + 3x^3 + x + 19
    Derivative = 16x^3 + 9x^2 + 1

    That’s all there is to taking the derivative of a polynomial!

    ---------------------------------------------------------------------------------------------


    Topic: Critical Points

    Aim: How to Determine Critical Points

    First, it is very important for math terms to be defined. Critical points are the points on the graph where the derivative, or slope, equals zero or does not exist. The points are called "CRITICAL POINTS" because they represent where something important takes place in the graph of the function.

    It is often useful to find critical points because all relative maxima and minima occur at critical points. However, the opposite is not true. It is possible for a critical point to not be a relative extrema.

    The typical procedure for finding critical points is to first find the derivative and then set it equal to zero. Find all points where the derivative is 0. Then, look for points where the derivative is undefined (often where a denominator equals 0).

    Example: Find the critical points of the function f(x) = x3 - 3x.

    Step 1) Find the derivative of each term separately using the power rule.

    d
    --- (x3) = 3(x2)
    dx

    Do likewise to the second term.

    d
    --- (-3x) = -1(3x0)
    dx

    d
    --- (-3x) = -3
    dx

    The derivative is f ' (x) = 3x2 - 3.

    Step 2) Set derivative to equal zero to find the x values of our critical points.

    3x2 - 3 = 0

    Solve for x by factoring out a 3:

    3(x2-1) = 0

    Divide both sides by 3:

    x2-1 = 0
    x^2-1 = 0 becomes (x+1) (x-1) = 0

    Solve for x individually.

    x + 1 = 0 and x - 1 = 0

    x = -1 and x = 1

    Step 3) Plug the values for x into the original function to find the y values that will make up our critical points.

    This is our original function: f(x) = x3 - 3x

    f(-1) = (-1)3 - 3(-1)

    f(-1) = 2

    So, our first critical point is (-1, 2).

    Next: do the same for x = 1.

    f(1) = (1)3 - 3(1)

    f(1) = -2

    So, our second critical point is (1, -2).

    To get a geometric view of the critical points, graph the original function.
    After graphing, you will notice that at point (-1, 2) the graph begins to descend and at point (1, -2) the graph begins to ascend:

    Dont mind the 8X's replace them with 8 X's
    Offline Profile Quote Post Goto Top
     
    Ramen Hood
    Paul's dumb Alex is cool

    Yeah, the clock would be a little pointless. Because on most peoples(including mine) computers, it says the time right there on the screen.
    Ron Paul in '08
    Posted Image
    <img src='http://www.flashflashrevolution.com/profile/ffrsiggy/Ramen Hood/0313.png' border='0' alt='user posted image'>
    Offline Profile Quote Post Goto Top
     
    APM
    Member Avatar


    Now where have I seen that lesson before...? :P
    Offline Profile Quote Post Goto Top
     
    UltraMegaGigaBowser
    Member Avatar
    Minnesota High School Hockey
    is it just me... or does anybody else wonder WHO IN THE WORLD WOULD WASTE TIME TYPING A MATH PROBLEM THAT LONG AND THEY KNO NOBODY WILL DO IT? It seems very wierd 2 me :dizzy: .


    Anyways i have a clock om my computer and anytime i need help i pm CTT or Matt. And i need help alot.

    UR 2 techinical ganondorf... this place is suppose to be fun!
    Offline Profile Quote Post Goto Top
     
    APM
    Member Avatar


    UltraMegaGigaBowser
    Apr 24 2005, 01:56 PM
    is it just me... or does anybody else wonder WHO IN THE WORLD WOULD WASTE TIME TYPING A MATH PROBLEM THAT LONG AND THEY KNO NOBODY WILL DO IT? It seems very wierd 2 me :dizzy: .


    Anyways i have a clock om my computer and anytime i need help i pm CTT or Matt. And i need help alot.

    UR 2 techinical ganondorf... this place is suppose to be fun!

    Why would you think math help sites are bad? They're not a waste of time at all...
    Offline Profile Quote Post Goto Top
     
    Ganondorf718
    Bombette
    -Yoshi-
    Apr 24 2005, 02:13 PM
    UltraMegaGigaBowser
    Apr 24 2005, 01:56 PM
    is it just me... or does anybody else wonder WHO IN THE WORLD WOULD WASTE TIME TYPING A MATH PROBLEM THAT LONG AND THEY KNO NOBODY WILL DO IT? It seems very wierd 2 me :dizzy: .


    Anyways i have a clock om my computer and anytime i need help i pm CTT or Matt.  And i need help alot.

    UR 2 techinical ganondorf... this place is suppose to be fun!

    Why would you think math help sites are bad? They're not a waste of time at all...

    Yeah, i did not type that whole thing... psh... i Crtl + C it into here... :D
    Offline Profile Quote Post Goto Top
     
    UltraMegaGigaBowser
    Member Avatar
    Minnesota High School Hockey
    Oh... OK!!! :woot:

    you have to admit tho... Math sites r pretty dull!!!
    Offline Profile Quote Post Goto Top
     
    1 user reading this topic (1 Guest and 0 Anonymous)
    « Previous Topic · Any ideas? · Next Topic »
    Add Reply
    • Pages:
    • 1
    • 2