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Math Help - I can't figure this interval out.

  1. #1
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    I can't figure this interval out.

    Let f(x)=6x+\frac{1}{x} find:
    f'(x)>0 for  x \in _____?
    f'(x)<0 for x \in _____?

    for the derivative I got 6-\frac{1}{x^2}.

    I set that to zero to find the critical points, and found \pm \sqrt{\frac{1}{6}}

    For f'(x)>0 I got (-\infty,-\sqrt{\frac{1}{6}}\ ] \cup \[ \sqrt{\frac{1}{6}},\infty )
    For f'(x)<0 I got [ -\sqrt{\frac{1}{6}},0 ) \cup (0,\sqrt{\frac{1}{6}} ]

    I was told that these are incorrect, and I don't understand why. I even graphed the derivative and this looked to me to be correct. Am I close? Please let me know where I am wrong. I really appreciate it. Thanks.
    Last edited by KhanDisciple; February 20th 2013 at 09:11 PM.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: I can't figure this interval out.

    perhaps because you used  (-\infty, -\sqrt{\frac{1}{6}}] \cup [\sqrt{\frac{1}{6}}, \infty)instead of  (-\infty, -\sqrt{\frac{1}{6}}) \cup (\sqrt{\frac{1}{6}}, \infty) (notice the ] replaced with ) ).at  \pm \sqrt{\frac{1}{6}} f'(x) is 0, it shouldnt be included as a value for which f'(x) > 0.

    Same reasoning for your interval for f'(x) < 0
    Last edited by jakncoke; February 20th 2013 at 09:45 PM.
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  3. #3
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    Re: I can't figure this interval out.

    Thanks a ton jakncoke, I would gladly buy you one given the chance.
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  4. #4
    Junior Member EliteAndoy's Avatar
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    Re: I can't figure this interval out.

    I think you missed one critical point, that is where f'(x) is undefined. which is when x=0. So if you evaluate the first derivative test. The function is decreasing at (0,sqrt(1/6)) and increasing at (sqrt(1/6),infinity) because remember that any value of x less than or equal to zero is now allowed to be used as a critical point. Try this one out and hope this helps!
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