mathcalculus
mathcalculus
HELP: Determine the values of where the tangent line is horizontal for the function:



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mathcalculus
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mathcalculus
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i know i have to find the derivative and set it to zero but it's not coming out to be correct. can anyone help me

mathcalculus
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so i did 3x^26x4=0

mathcalculus
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i tried to use the quadratic formula and got sqrt(21)+3/3

mathcalculus
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+ or  sqrt(21)+3/3

mathcalculus
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also aka 2.52753...

dumbcow
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derivative seems correct

mathcalculus
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to find what x = to... ?

mathcalculus
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i used quadratic formula

mathcalculus
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@dumbcow dont i have to do that to find x??

mathcalculus
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im getting it wrong,

dumbcow
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yes everything looks correct to me...do they want exact or decimal answer?


mathcalculus
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can i use quadratic formula for ths one?

mathcalculus
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@dumbcow

mathcalculus
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??

mathcalculus
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????????????????????????

dumbcow
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yes of course you use quadratic formula...thats how you got your answer that i said looked correct

mathcalculus
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im not getting that using quadratic formula

mathcalculus
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im getting something totally different and wrong

dumbcow
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\[x = \frac{3 \pm \sqrt{21}}{3}\]

mathcalculus
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yes! but from 6+2*sqrt(21)/6?

mathcalculus
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shouldnt it be 1+ or  sqrt 21/3

dumbcow
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same thing, just a matter of splitting up fraction or keeping it as single fraction

mathcalculus
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no it's not because the answer is 1+ sqrt (7/3) gives me 2.52753 and 1+sqrt 21/3 gives me 1.86086!

mathcalculus
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@dumbcow

dumbcow
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hmm
\[\frac{\sqrt{7}}{\sqrt{3}} = \frac{\sqrt{21}}{3} \]

mathcalculus
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hmmm?

mathcalculus
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@dumbcow

mathcalculus
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try it on a calculator.


dumbcow
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oh you added the "1" before dividing by "3"

mathcalculus
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okay from my answer.... not the internet.. 6+2 sort(21)/3.... where do i go from there?

dumbcow
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well it should be a 6 on bottom

mathcalculus
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no i didn't add anything.

mathcalculus
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okay sure a 6. i thought it could be a 3 to narrow it down

mathcalculus
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so 6+or2sqrt(21)/6

dumbcow
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i dont see the problem, you posted answer in beginning...
" i tried to use the quadratic formula and got sqrt(21)+3/3 "

mathcalculus
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so it is a 3 in the denominator!

dumbcow
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\[\frac{6 \pm 2\sqrt{21}}{6} = \frac{2(3 \pm \sqrt{21})}{6} = \frac{3 \pm \sqrt{21}}{3}\]