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Math Help - System of Equations (Check my work)

  1. #1
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    System of Equations (Check my work)

    So I've been doing this problem for a while now and I keep getting the wrong answer:

    Find a cubic function, in the form below, that has a local maximum value of 4 at -4 and a local minimum value of 0 at 2.

    f(x)=ax^3+bx^2+cx+d

    So I just use can obtain 4 equations by subsituting (-4,4) and (2,0) into f(x) and into the derivative of f(x) where f'(x)=3ax^2+2bx+c=0:

    For f(x): at x=-4, f(x)=-64a+16b-4c+d-4=0
    at x=2, f(x)=8a+4b+2c+d=0
    For f'(x): at x=-4, f'(x)=48a-8b+c=0
    at x=2, f'(x)=12a+4b+c=0

    So when I solve it I get a= -1/93, b= -1/31, c= 8/31, d= -28/93 but that only solves 3 of the above equations...so I'm doing something wrong. I'll post all my work in the next post since it's a bit long. Also, I don't know how to use matrices so try to refrain from using it. Thanks.
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  2. #2
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    Quote Originally Posted by VectorRun View Post
    So I've been doing this problem for a while now and I keep getting the wrong answer:

    Find a cubic function, in the form below, that has a local maximum value of 4 at -4 and a local minimum value of 0 at 2.

    f(x)=ax^3+bx^2+cx+d
    I don't know how to use matrices so try to refrain from using it.
    then you better learn ... calculators these days actually do matrix operations. If you insist on doing grunt work to solve for a 4 equation system, be my guest.

    a = 1/27
    b = 1/9
    c = -8/9
    d = 28/27
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  3. #3
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    Step 1: 48a=8b-c, a=\frac {8b-c}{48}
    So input that into the rest:

    12\frac {8b-c}{48}+4b+c=0, \frac{24b+3c}{4}=0
    -64\frac {8b-c}{48}+16b-4c+d-4=0, \frac {16b+52c+3d-12}{3}=0
    8\frac {8b-c}{48}+4b+2c+d=0, \frac{32b+11c+6d}{6}

    Step 2: \frac{24+3c}{4}=0, b=\frac {-1}{8}c
    Plug that into the the remaining two:

    \frac{16(\frac{-1}{8}c)+52c+3d-12}{3}=0, \frac{50c+3d-12}{3}=0
    \frac{32(\frac{-1}{8}c)+11c+6d}{6}=0, \frac{7c+6d}{6}=0

    Step 3: c=\frac{12-3d}{50}

    \frac{7\frac{12-3d}{50}+6d}{4}=0, \frac {84+279d}{200}=0, d=\frac{-28}{93}

    Then working back from there I got the rest I mentioned but it doesn't seem to work. Also, if you can get the answer in matrices I don't really mind but critique on this work would be preferred.
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  4. #4
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    We haven't been taught matrices yet but my higher-year friend said that you can use it solve them. I'll learn it eventually but I'm technically not supposed to use matrices to solve this problem.

    Wow, I sure did something wrong...Gah, better go check again.
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  5. #5
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    Okay, I redid the whole question and finally got the answer the long way. I don't know why but I sure complicated myself...I should've just subtracted the similar ones from each other since that would've gotten rid of the "d"s in ones relative to f(x). Haha, I need to simplify more often. Anyways, thanks for all the help. I can't wait to learn matrices but that's actually going to happen in my data management class next semester. Bleh, such a long wait! =/
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