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anonymous
 3 years ago
find the critical points: (attached)
anonymous
 3 years ago
find the critical points: (attached)

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anonymous
 3 years 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

anonymous
 3 years 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can someone tell me what i did wrong?

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

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

zepdrix
 3 years 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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do we have to use quadratic formula? i just factored

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

anonymous
 3 years 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
 3 years 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.

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

zepdrix
 3 years 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 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh okay thank you @zepdrix :)!

anonymous
 3 years 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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