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Homework Help: First and second derivatives

  1. Nov 1, 2008 #1
    This problem is fairly straightforward. I have a function and domain:
    [tex]f(x)=2ln(x^{2}+3)-x -3\leqx\leq5[/tex]
    The question eventually asks for the inflection points, which I know occur when f''(x) is zero or undefined.
    I found the first derivative and second derivative to be, respectively:
    [tex]f'(x)=(4x)(x^{2}+3)^{-1}-1[/tex]
    [tex]f"(x)=(4)(x^{2}+3)^{-1}-(4x)(x^{2}+3)^{-2}[/tex]
    I have double checked my math and can't seem to find anything wrong with it.
    Also, I have looked at a calculator and it shows that the original function does have two inflection points, but when I try to set f"(x) to zero or to be undefined (devide by zero), I always end up with an impossible solution. I don't beleive that this involves irrational numbers, so I appreciate any help at all.:confused:
     
  2. jcsd
  3. Nov 1, 2008 #2

    HallsofIvy

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    Science Advisor

    Your second derivative is wrong.

    You have
    [tex]f"(x)=(4)(x^2+3)^{-1}-(4x)(x^2+3)^{-2}[/tex]

    With [itex]f'(x)= (4x)(x^2+3)^{-1}- 1[/itex] the second derivative is
    [tex]4(x^2+ 3)^{-1}- (4x)(x^2+ 3)^{-2}(2x)= 4(x^2+ 3)^{-1}- 8x^2(x^2+3)^{-2}[/tex]
    [tex]= \frac{4(x^2+ 3)}{(x^2+ 3)^{-2}}- \frac{8x^2}{(x^2+ 3)^{-2}}= \frac{-8x^2+ 3}{(x^2+ 3)^{-2}}[/tex]
    The second derivative always exists but there are two places where it is 0.
     
  4. Nov 2, 2008 #3
    Thank you very much, I see it now.
    I figured it was something in the calculation of the first to second, I just couldn't see it.
     
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