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Math Help - Help! sketching 3rd deg. equation?

  1. #1
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    Help! sketching 3rd deg. equation?

    ** This is seriously my 7th time creating this thread, please excuse my hastiness**

    Hey there everyone, I would love for someone to help me out with this problem ASAP. Ive looked through my math book for the sugested method to do this based on example but I couldn't find it.

    Sketch f(x), identify coordinates of any local min/max.

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

    **What I did:: Well, since it's a Positive 3rd deg. there will be 2 turning points at most.. so.. in the derivative of f(x) those turning points should show up as zeros of the derivative?

    f^1(x) = 1-6x+3x

    quadratic formula:
    0 = [6+/- (root) 24]/6
    x= 1.82 ; x= 0.2

    **soo.. this is where im stuck..
    do i plug the found values of x back into f(x)?
    if so: f(0.2) = 2-3(0.2)^2+(0.2)^3
    f(0.2) = 1.88

    ***Does this mean one of the local min/max coords is (0.2, 1.88)?

    Thanks in advance to anyone who dares to tackle this monster
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  2. #2
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    Joined
    Nov 2008
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    Hello buddy

    Quote Originally Posted by Lucidliving View Post
    ** This is seriously my 7th time creating this thread, please excuse my hastiness**

    Hey there everyone, I would love for someone to help me out with this problem ASAP. Ive looked through my math book for the sugested method to do this based on example but I couldn't find it.

    Sketch f(x), identify coordinates of any local min/max.

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

    **What I did:: Well, since it's a Positive 3rd deg. there will be 2 turning points at most.. so.. in the derivative of f(x) those turning points should show up as zeros of the derivative?
    Sounds good.

    Quote Originally Posted by Lucidliving View Post
    f^1(x) = 1-6x+3x
    Where does the '1' come from?

    f'(x) = 3x^2-6x

    because the derivative of 2 is 0;
    Quote Originally Posted by Lucidliving View Post
    quadratic formula:
    0 = [6+/- (root) 24]/6
    x= 1.82 ; x= 0.2
    That's wrong anyway

    Consider
    f'(x) = 3x^2-6x  = x(3x-6) = 0

    \Rightarrow x = 0 and x = 2


    Quote Originally Posted by Lucidliving View Post
    **soo.. this is where im stuck..
    do i plug the found values of x back into f(x)?
    if so: f(0.2) = 2-3(0.2)^2+(0.2)^3
    f(0.2) = 1.88
    Yes, you do. Because you want to find the x (you already know them) and y-coordinates.
    Quote Originally Posted by Lucidliving View Post
    ***Does this mean one of the local min/max coords is (0.2, 1.88)?

    Thanks in advance to anyone who dares to tackle this monster
    You have to find it out by considering the second derivative at x = 0 and x=2

    Is
    f''(0) > 0 => min

    f'(0) < 0 => max

    f''(2) > 0 => min

    f''(2) < 0 => max

    Yours
    Rapha
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  3. #3
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    Thank you so much! !!!!
    !!!
    !!!!!

    Im so glad there are people like you willing to help for absolutely no payment of any kind. Except perhaps a social incentive.

    I wish I could give back.. but thats why Im here, I cant.. =/
    Unless someone wants to know about Physics or Economics!
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  4. #4
    MHF Contributor Amer's Avatar
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    about physics there is a forum about physics

    http://www.physicshelpforum.com/physics-help/




    see this link

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