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anonymous

  • 5 years ago

f(x)=(3+9ln(x))/x i need to find the x coordinate of the absolute max...

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  1. amistre64
    • 5 years ago
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    then lets get the derivative :)

  2. anonymous
    • 5 years ago
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    \[-9\ln x-6/x^2\] i think this is it

  3. anonymous
    • 5 years ago
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    u need help

  4. amistre64
    • 5 years ago
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    (x)(9/x) - (3+9ln(x))(1) ------------------- x^2 3(2 -3ln(x)) ----------- x^2

  5. anonymous
    • 5 years ago
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    help me

  6. amistre64
    • 5 years ago
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    I need help? like professional help?

  7. anonymous
    • 5 years ago
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    lol

  8. amistre64
    • 5 years ago
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    .....me and my voices are content :)

  9. anonymous
    • 5 years ago
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    ok so now we need to find the absolute max by setting f' to 0

  10. amistre64
    • 5 years ago
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    0 will give us critical points, that we can test for minand max

  11. anonymous
    • 5 years ago
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    Do you have a particular bound in which you're looking for the max, or is it over (-infinity, +infinity)

  12. amistre64
    • 5 years ago
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    polpak is also good at this stuff :)

  13. anonymous
    • 5 years ago
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    x>0

  14. anonymous
    • 5 years ago
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    Well.. What is the limit as x approaches infinity is another thing to consider then when looking for absolute max. You have to check for critical points, but also end points. If the function goes to infinity as x goes to infinity then it has no absolute max value.

  15. amistre64
    • 5 years ago
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    if I see it right: 2 = 3ln(x) is what we are looking to solve

  16. anonymous
    • 5 years ago
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    yes that was correct...lol

  17. amistre64
    • 5 years ago
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    cbrt(e^2) is a possible critical point.... the only one I can come up with

  18. amistre64
    • 5 years ago
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    as x gets large, this goes to zero I think

  19. anonymous
    • 5 years ago
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    Yes, it does.

  20. amistre64
    • 5 years ago
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    the 2nd derivative I believe is: 3(7+6ln(x)) --------- x^3

  21. anonymous
    • 5 years ago
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    x=e^(3/2) would be the max value. As it approaches 0 as x goes to zero or infinity.

  22. anonymous
    • 5 years ago
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    i finished that problem i got a different one.... mean slope=-10 f'x=6x^2-12x-90 mean value theroem

  23. amistre64
    • 5 years ago
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    at cbrt(e^2) the 2nd derivative would be 33/e^2; which is positive.... I believe that it is a MIN at that point

  24. amistre64
    • 5 years ago
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    finished? kaputs? done? howd we do :)

  25. anonymous
    • 5 years ago
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    the answer was x=e^(2/3)

  26. amistre64
    • 5 years ago
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    hmmm.... I was right :) I like being right. even when im wrong...

  27. amistre64
    • 5 years ago
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    whats that new problem? cant quite decipher it

  28. anonymous
    • 5 years ago
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    Ok it was like another we did the average or mean slope =-10 and my f'x=6x^2-12x-90 I need to find values that make this equal to the slope

  29. amistre64
    • 5 years ago
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    let find the derivative that equals that slope to get the equation for the tangent line from it :)

  30. anonymous
    • 5 years ago
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    That is the derivative...

  31. amistre64
    • 5 years ago
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    ohh.... my eyes are going :) make it equal to -10; then add 10 to both sides to get it equal to 0 again and sove for x right?

  32. amistre64
    • 5 years ago
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    6x^2-12x-90 = -10 6x^2 -12x -80 = 0 6(x^2 -2x - 40/3) nice numbers :) remember to stop me when I do something stupid :)

  33. amistre64
    • 5 years ago
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    do quad formula on it to speed things up :) 144-(4)(-80)(6) 144 + 1920 12 +- sqrt(2064) --------------- right? 12

  34. amistre64
    • 5 years ago
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    1 +- sqrt(129)/3

  35. amistre64
    • 5 years ago
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    since the original will be a cubic, it stands to reason that there are 2 places that give us a slope of -10

  36. amistre64
    • 5 years ago
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    in fact, we can just suit up f'(x) to get that function right?

  37. amistre64
    • 5 years ago
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    (S) 6x^2-12x-90 dx 2x^3 -6x^2 -90x +c is in the family of curves

  38. amistre64
    • 5 years ago
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    we can drop the constant if we want to just use the x values to get it... right?

  39. anonymous
    • 5 years ago
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    \[2(3x^2 -6x -40) = 0\] \[\rightarrow 3x -6x -40 = 0\] \[\rightarrow x = \frac{6 \pm \sqrt{516}}{6}\]

  40. anonymous
    • 5 years ago
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    So at those x values the slope is -10.

  41. amistre64
    • 5 years ago
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    should be.... if I didnt mess it up :)

  42. anonymous
    • 5 years ago
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    Well I did it myself, so if your x matches mine it's probably right.

  43. amistre64
    • 5 years ago
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    I see the 1 :) can 516 get reduced?

  44. amistre64
    • 5 years ago
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    maybe to 129?

  45. amistre64
    • 5 years ago
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    close.... I mighta messed something up tho :)

  46. anonymous
    • 5 years ago
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    The equation is: \[f(x)=\frac{3+9 \text{Log}(x)}{x}\] The first derivative, \[f'(x)=\frac{9}{x^2}-\frac{3+9 \text{Log}(x)}{x^2} \] simplified, is: \[f'(x)=\frac{6-9 \text{Log}(x)}{x^2} \] Set the Numerator to zero and solve for x \[6-9 \text{Log}(x)=0\] \[x\text{=}e^{2/3} \] Pluggin x into f(x) gives the coordinates of the maximum funtion value: \[\left\{e^{2/3},\frac{9}{e^{2/3}}\right\}\text{ = }\{1.94773,4.62075\} \] Refer to the attachment plot of f(x).

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