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Math Help - concavity

  1. #1
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    concavity

    The graph of the function y=x^3 + 6x^2 + 7x - 2cosx changes concavity at x=....?

    We're supposed to use a calculator and the answer is -1.89 but how do you find that?
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  2. #2
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    Find the 2nd derivative, set to 0 and solve for x.

    \frac{d^{2}}{dx^{2}}[x^{3}+6x^{2}+7x-2cos(x)]=2cos(x)+6x+12

    Solve 2cos(x)+6x+12=0 for x.
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  3. #3
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    Quote Originally Posted by abcocoa View Post
    The graph of the function y=x^3 + 6x^2 + 7x - 2cosx changes concavity at x=....?

    We're supposed to use a calculator and the answer is -1.89 but how do you find that?
    y=x^3+6x^2+7x-2\cos x
    y'=3x^2+12x+7+2\sin x
    y''=6x+12+2\cos x
    Solve,
    2\cos x+6x+12=0
    \cos x+3x+4=0

    You need to use Newton's method.
    Graph the function and see approximate value. Shown Below.
    We need to find zero for the function,
    f(x)=\cos x+3x+4
    f'(x)=-\sin x+3
    We note that,
    x\approx -1
    Define the iterative sequence,
    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}
    x_{n+1}=x_n-\frac{\cos x_n+3x_n+4}{3-\sin x_n}
    x_{n+1}=\frac{3x_n-x_n\sin x_n-\cos x_n-3x_n-4}{3-\sin x_n}
    x_{n+1}=\frac{x_n\sin x_n+\cos x_n+4}{\sin x_n-3}
    The the iterative sequnce is,
    -1,-1.401,-1.3925,-1.3925,-1.3925,...
    Thus, the solution is approximately,
    x\approx -1.3925
    Attached Thumbnails Attached Thumbnails concavity-picture4.gif  
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  4. #4
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    but....

    the answer is -1.89 not -1.39...

    Also my problem is solving 6x+12+2cosx=0. How do i solve this?
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  5. #5
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    Quote Originally Posted by abcocoa View Post
    the answer is -1.89 not -1.39...
    The answer you gave it wrong, just substituet it into the equation and see that it is not zero.
    Also my problem is solving 6x+12+2cosx=0. How do i solve this?
    I posted the solution above.
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  6. #6
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    Smile

    PH, I think you have a small typo in your derivative. Should be

    cos(x)+3x+6. You have cos(x)+3x+4

    The solution is, indeed, -1.89. I ran it through the ol' TI-92 to make doubly sure.

    You may use the Intermediate Value Theorem. You can try x=-1.88 and get 0.055699

    x=-1.9 and get -0.0232895.

    Note the sign change. Therefore, you know it's between -1.88 and -1.9. Whittle away.
    Last edited by galactus; January 7th 2007 at 05:22 PM.
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