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mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i found the derivative then i set it to zero. i factored then found the critical points: x=8/3

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i tried to find y by plugging in the x into original function.... then got this: (8/3,419.92) and (4/5,58.16)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0can someone tell me what i did wrong?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0critical points *** x=8/3 and x=4/5

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So did you get this for your derivative function?\[\large h'(x)=15x^2116x32\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Setting 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
 one year ago
Best ResponseYou've already chosen the best response.1I'm coming up with different solutions, lemme know if that setup looks like what you have.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0do we have to use quadratic formula? i just factored

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You were able to factor it into fractions? lol... You must've factored something incorrectly :[

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.1Mmm yah something like that, lemme try to factor it real quick. I'm not very good with factoring myself :) lol.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i got: x=8 and x=4/15 by using the quadratic formula

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Yah 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 :)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0oh okay thank you @zepdrix :)!

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i think we can't factor so we use quadratic right away..
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