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Math Help - local man, local min, inflection point problem

  1. #1
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    local man, local min, inflection point problem

    find a,b,c, and d such that the cubic equation f(x)=ax^3+bx^2+cx+d has a local maximum at (3,3); a local minimum at (5,1); and an inflection point at (4,2).

    f'(x)=3ax^2+2bx+c
    f''(x)=6ax+2b

    local min and local max f'(x)=0 and inflection point f''(x)=0

    so plugging in x=3 and x=5 for max and min into f'(x) and setting it equal to 0 i get
    27a+6b+c=0
    75a+10b+c=0
    plugging in x=4 for f''(x) and setting it equal to 0 i get
    24a+2b=0

    also i know i have to plug in x=3, x=5, and x=4 into the original equation while setting each to 3, 1, and 2 to get
    27a+9b+3c+d=3
    125a+25b+5c+d=1
    64a+16b+4c+d=2

    but i know have a system of 6 equations with 4 unknowns and i know this system wont give me the values i am looking for since i have more equations than unknowns
    however my professor did mention that i could get rid of 2 of the equations so ill have 4 equations with 4 unknowns, the reason why i am stuck up to this point is that i do not know which two equations to get rid of
    i want to choose the the two equations dealing with the inflection point, that is
    64a+16b+4c+d=2
    24a+2b=0
    but the only reason i can think of for choosing these two is because since there is a max at x=3 and min at x=5 then the graph has to increase hit the max at x=3 then decrease till it hits the min, so in between x=3 and x=5 the graph has to change concavity, and the question already says this happens at x=4 so we can just ignore those 2 equations, however im not too sure of my reasoning there haha
    also could i just get rid of any 2 equations and still get the right answer?

    please help and sorry for making this such a long post, just wanted anyone who sees my post what ive done so far and more

    thanks in advance
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    All your equations are right. Now, for example

    Quote Originally Posted by Miguel View Post
    27a+6b+c=0
    75a+10b+c=0
    24a+2b=0

    We have b = - 12 a . Substitute in the first and second equation and you'll obtain:

    - 45 a + c = 0
    - 45 a + c = 0

    So, those three equations are equivalent to

    12 a + b = 0
    45 a - c = 0

    Hope this helps.
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  3. #3
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    ohhhh i see, wow i knew it was going to be something very easy
    well anyways thanks a lot for the helpful hint
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