anonymous
  • anonymous
find the critical points: (attached)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
i found the derivative then i set it to zero. i factored then found the critical points: x=-8/3
anonymous
  • anonymous
correct?

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anonymous
  • anonymous
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)
anonymous
  • anonymous
can someone tell me what i did wrong?
anonymous
  • anonymous
critical points *** x=-8/3 and x=4/5
zepdrix
  • zepdrix
So did you get this for your derivative function?\[\large h'(x)=15x^2-116x-32\]
anonymous
  • anonymous
yes
zepdrix
  • zepdrix
Setting it equal to zero, tossing it into the Quadratic Formula gives us,\[\large x=\frac{116 \pm \sqrt{116^2+4\cdot15\cdot32}}{2\cdot15}\]
zepdrix
  • zepdrix
I'm coming up with different solutions, lemme know if that setup looks like what you have.
anonymous
  • anonymous
do we have to use quadratic formula? i just factored
zepdrix
  • zepdrix
You were able to factor it into fractions? lol... You must've factored something incorrectly :[
anonymous
  • anonymous
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?
zepdrix
  • zepdrix
Mmm yah something like that, lemme try to factor it real quick. I'm not very good with factoring myself :) lol.
anonymous
  • anonymous
lol ok
anonymous
  • anonymous
i got: x=8 and x=-4/15 by using the quadratic formula
zepdrix
  • zepdrix
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 :)
anonymous
  • anonymous
oh okay thank you @zepdrix :)!
anonymous
  • anonymous
i think we can't factor so we use quadratic right away..

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