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anonymous

  • one year ago

Need Help! Use f(x) = (lnx) / Sqrtx to do the following below: a) find where its increasing or decreasing b) find local max and mins values c) find intervals of concavity and inflection points

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  1. Zale101
    • one year ago
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    Have you started? If so, then jot everything down.

  2. Zale101
    • one year ago
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    Start by the first derivative test.

  3. UsukiDoll
    • one year ago
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    \[f(x) = \frac{lnx}{\sqrt{x}}\]

  4. Zale101
    • one year ago
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    Use quotient rule

  5. UsukiDoll
    • one year ago
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    and this time I realized my error from earlier.... just setting that f(x) = 0 is equilibrium which is not what we need so obtain the first derivative and set f'(x) = 0 to solve for critical points then pick a value between the intervals and plug it into f'(x) to see if it increases or decreases you will need to obtain the second derivative for concavity f''(x) = 0 and repeat the steps above to see if we have concave up or concave down.

  6. Zale101
    • one year ago
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    +1 Usukidoll Also, please post all your work. Posting just the question is not telling us much on what you need help with.

  7. UsukiDoll
    • one year ago
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    @Zale101 that's what happens when insomnia for weeks plagues my brain... the thinking goes nuts @______@

  8. Zale101
    • one year ago
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    I understand. It makes it hard for tutors to tutor properly when they dont know what the real problem the tutee is asking. Do they want us to explain everything in the concept? Or is there some certain part of the problem they are confused with?

  9. anonymous
    • one year ago
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    Oops, didnt see these comments until now. Either way, I also know how to get the first derivative of this

  10. anonymous
    • one year ago
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    which is: sqrt of x * (1/x) - (lnx) (1/2x^-1/2) all over (sqrt of x) ^2

  11. anonymous
    • one year ago
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    But I dont know how to go about from there

  12. ChillOut
    • one year ago
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    Find the stationary point, I.E, f'(x)=0. Then check for adjacent points to see if the derivative is increasing or deacreasing nearby.

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