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anonymous

  • one year ago

Find the critical numbers of f(x) = x^(4/5)(7x-28)^2

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  1. anonymous
    • one year ago
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    \[f(x)=x^{\frac{4}{5}}(7x-28)^2\] right?

  2. anonymous
    • one year ago
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    Yep! :)

  3. anonymous
    • one year ago
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    I know I have to find where the derivative = 0 or does not exist... Things just get a little bit messy when I derive it

  4. anonymous
    • one year ago
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    it is a huge mess, that is why

  5. anonymous
    • one year ago
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    Lol XD

  6. anonymous
    • one year ago
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    one place it will not exist is at \(x=0\) since the derivative of \(x^{\frac{4}{5}}\) is \[\frac{4}{5}x^{-\frac{1}{5}}=\frac{4}{5}\frac{1}{\sqrt[5]{x}}\]

  7. anonymous
    • one year ago
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    Ohh right!

  8. anonymous
    • one year ago
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    you gotta use the chain rule and product rule to find the derivative or else expand the whole thing out and use the power rule repeatedly you still have to add up the fractions in any case

  9. anonymous
    • one year ago
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    it is a raft of boring algebra to get the derivative as one fraction i could walk you through it probably, but best idea is to cheat

  10. anonymous
    • one year ago
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    turns out the numerator is \[98(7x^2-36x+32)\] and you can set that equal to zeros and solve via the quadratic formula

  11. anonymous
    • one year ago
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    Yep I used chain rule and product rule and I got this... \[f'(x) = \frac{ 4 }{ 5 }x ^{-1/5}(7x-28)^2+14x ^{4/5}(7x-28)\]

  12. anonymous
    • one year ago
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    nope

  13. anonymous
    • one year ago
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    Oh how did you get that?

  14. anonymous
    • one year ago
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    oh wait how did you get the second part?

  15. anonymous
    • one year ago
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    yeah i think that is wrong

  16. anonymous
    • one year ago
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    Isn't it product rule?

  17. anonymous
    • one year ago
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    oh no it isn't my mistake you are right

  18. anonymous
    • one year ago
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    now here is what you have to do

  19. anonymous
    • one year ago
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    in order to write it as one fraction, (since the first part is a fraction) multiply top and bottom by \(\sqrt[5]{x}\)

  20. anonymous
    • one year ago
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    Oh so we multiply it on the second piece only?

  21. anonymous
    • one year ago
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    rational exponents are your friend when it comes to taking derivatives, but not when it comes to finding critical points or actually evaluating the derivative you need to get it out of exponential notation and write it as radicals

  22. anonymous
    • one year ago
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    \[\frac{4(7x-28)^2}{5\sqrt[5]{x}}+14\sqrt[5]{x^2}(7x-28)\] is what you are looking at

  23. anonymous
    • one year ago
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    to put it over the same denominator multiply the second part top and bottom by \(\sqrt[5]{x}\) which is nice because the radical will go away

  24. anonymous
    • one year ago
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    the numerator will be \[4(7x-28)^2+14x(7x-28)\] which you can do the algebra with to get a quadratic to solve

  25. anonymous
    • one year ago
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    Ahh I see!! So all I have to do now, is expand the numerator and set it = 0 and those are gonna be my critical numbers right?

  26. anonymous
    • one year ago
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    damn i made a typo, but you can probably spot it

  27. anonymous
    • one year ago
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    Hmmm... where?

  28. jim_thompson5910
    • one year ago
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    This is what I'm getting \[\large f(x) = x^{4/5}(7x-28)^2\] \[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+14x^{4/5}(7x-28)\] \[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+14x^{4/5}(7x-28) {\color{red}{\times \frac{5x^{1/5}}{5x^{1/5}}}}\] \[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+\frac{70x(7x-28)}{5x^{1/5}}\] \[\large f \ '(x) = \frac{4(7x-28)^2+70x(7x-28)}{5x^{1/5}}\] \[\large f \ '(x) = \frac{98(x-4)(7x-8)}{5x^{1/5}}\]

  29. anonymous
    • one year ago
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    Oh the 5 was missing!! Thanks so much @jim_thompson5910 ! :)

  30. jim_thompson5910
    • one year ago
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    no problem

  31. anonymous
    • one year ago
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    Expanding it, I get huge numbers, is that how it's supposed to be??

  32. jim_thompson5910
    • one year ago
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    set it equal to 0 and solve for x

  33. jim_thompson5910
    • one year ago
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    namely the numerator

  34. anonymous
    • one year ago
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    How did you get the last line @jim_thompson5910 ?

  35. anonymous
    • one year ago
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    lol i was missing the five too doh!

  36. anonymous
    • one year ago
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    also os has be acting up, i kept trying to send the wolfram line http://www.wolframalpha.com/input/?i=x^%284%2F5%29%287x-28%29^2

  37. jim_thompson5910
    • one year ago
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    4(7x-28)^2 + 70x(7x-28) 2(7x-28)*[2(7x-28) + 35x] ... factor out 2(7x-28) 2(7x-28)*(14x-56 + 35x) 2(7x-28)*(49x - 56) 2[7(x-4)]*[7(7x-8)] 98(x-4)(7x-8)

  38. anonymous
    • one year ago
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    Ohhh ok! Yeah I was confused on how to simplify that! XD Geee lol

  39. anonymous
    • one year ago
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    Then I have to simplify the 2 parentheses right?

  40. jim_thompson5910
    • one year ago
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    yeah set it equal to 0 and use the zero product property

  41. anonymous
    • one year ago
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    Ok im gonna give that a try! :D

  42. anonymous
    • one year ago
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    Thanks so much @jim_thompson5910 and @satellite73 !!! :D

  43. jim_thompson5910
    • one year ago
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    you're welcome

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