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mathcalculus Group Title

find the critical points: (attached)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
  2. mathcalculus Group Title
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    i found the derivative then i set it to zero. i factored then found the critical points: x=-8/3

    • one year ago
  3. mathcalculus Group Title
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    correct?

    • one year ago
  4. mathcalculus Group Title
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    i tried to find y by plugging in the x into original function.... then got this: (-8/3,-419.92) and (4/5,-58.16)

    • one year ago
  5. mathcalculus Group Title
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    can someone tell me what i did wrong?

    • one year ago
  6. mathcalculus Group Title
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    critical points *** x=-8/3 and x=4/5

    • one year ago
  7. zepdrix Group Title
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    So did you get this for your derivative function?\[\large h'(x)=15x^2-116x-32\]

    • one year ago
  8. mathcalculus Group Title
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    yes

    • one year ago
  9. zepdrix Group Title
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    Setting it equal to zero, tossing it into the Quadratic Formula gives us,\[\large x=\frac{116 \pm \sqrt{116^2+4\cdot15\cdot32}}{2\cdot15}\]

    • one year ago
  10. zepdrix Group Title
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    I'm coming up with different solutions, lemme know if that setup looks like what you have.

    • one year ago
  11. mathcalculus Group Title
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    do we have to use quadratic formula? i just factored

    • one year ago
  12. zepdrix Group Title
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    You were able to factor it into fractions? lol... You must've factored something incorrectly :[

    • one year ago
  13. mathcalculus Group Title
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    i don't know i was told you can break it down bby multiplying the coefficient and constant= # and finding the factors and sums of that number. right?

    • one year ago
  14. zepdrix Group Title
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    Mmm yah something like that, lemme try to factor it real quick. I'm not very good with factoring myself :) lol.

    • one year ago
  15. mathcalculus Group Title
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    lol ok

    • one year ago
  16. mathcalculus Group Title
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    i got: x=8 and x=-4/15 by using the quadratic formula

    • one year ago
  17. zepdrix Group Title
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    Yah me too. \[\large 15=5\cdot3 \qquad \qquad -32=-4\cdot8\] That relates to your factoring method somehow..\[\large \frac{-4}{15}\qquad=\qquad \frac{-4}{3 \cdot 5}\] I just can't seem to figure out how :)

    • one year ago
  18. mathcalculus Group Title
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    oh okay thank you @zepdrix :)!

    • one year ago
  19. mathcalculus Group Title
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    i think we can't factor so we use quadratic right away..

    • one year ago
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