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Math Help - Find function with these minimums and maximums

  1. #1
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    Find function with these minimums and maximums

    Find values of a,b,c and d such that g(x) = ax^3+bx^2+cx+d has a local maximum of (2,4) and a local minimum at (0,0).

    How do I do this?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    Find values of a,b,c and d such that g(x) = ax^3+bx^2+cx+d has a local maximum of (2,4) and a local minimum at (0,0).

    How do I do this?
    You know four things g(0)=0,g(2)=4,g'(2)=0,g'(0)=0....haha and two more f''(2)<0...and f''(0)>0
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    Yes, that is the question...and what to do from there to find the original equation?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Uh...

    Quote Originally Posted by theowne View Post
    Yes, that is the question...and what to do from there to find the original equation?
    ...hmm four equations with four variables...hmm that sounds solvable...hmm...set it up =D haha...
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    Set what up? What four equations? The only equations I can see involved are the function, derivative and second derivative. I am attempting to learn this material, it'd be more helpful to see an example to learn from.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by theowne View Post
    Set what up? What four equations? The only equations I can see involved are the function, derivative and second derivative. I am attempting to learn this material, it'd be more helpful to see an example to learn from.
    f'(x)=3ax^2+2bx+c...we know that f'(0)=0...therfore 0=3a0^2+2b0+c,c=0...now you also know that f'(2)=0...so 0=3ax^2+2bx...now that you get the jist..set up your other equations and solve for the coefficents
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by theowne View Post
    Find values of a,b,c and d such that g(x) = ax^3+bx^2+cx+d has a local maximum of (2,4) and a local minimum at (0,0).

    How do I do this?
    f(x)=ax^3+bx^2+cx+d \mbox{ and } f'(x)=3ax^2+2bx+c

    With the above info we get

    using the point (0,0)

    f(0)=0=a(0)^3+b(0)^2+c(0)+d \iff d=0

    f'(0)=0=3a(0)^2+2b(0)+c \iff c=0

    since we know that c=d=0 we get

    f(x)=ax^3+bx^2 \mbox{ and } f'(x)=3ax^2+2bx

    using the point (2,4)

    f(2)=4=a(2)^3+b(2)^2 \iff 4=8a+4b

    f'(2)=0=3a(2)^2+2b(2) \iff 0=12a+4b \iff b=-3a

    solving this system by substition (or elimination)

     4 =8a+4(-3a) \iff 4 =-4a \iff a=-1

    so then b=3

    Finally we get

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