mathcalculus HELP: Determine the values of where the tangent line is horizontal for the function:
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
so i did 3x^2-6x-4=0
i tried to use the quadratic formula and got sqrt(21)+3/3
+ or - sqrt(21)+3/3
also aka 2.52753...
derivative seems correct
to find what x = to... ?
i used quadratic formula
@dumbcow dont i have to do that to find x??
im getting it wrong,
yes everything looks correct to me...do they want exact or decimal answer?
can i use quadratic formula for ths one?
@dumbcow
??
????????????????????????
yes of course you use quadratic formula...thats how you got your answer that i said looked correct
im not getting that using quadratic formula
im getting something totally different and wrong
\[x = \frac{3 \pm \sqrt{21}}{3}\]
yes! but from 6+2*sqrt(21)/6?
shouldnt it be 1+ or - sqrt 21/3
same thing, just a matter of splitting up fraction or keeping it as single fraction
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!
@dumbcow
hmm \[\frac{\sqrt{7}}{\sqrt{3}} = \frac{\sqrt{21}}{3} \]
hmmm?
@dumbcow
try it on a calculator.
http://www.wolframalpha.com/input/?i=1%2Bsqrt%287%2F3%29+%2C+1%2Bsqrt%2821%29%2F3
oh you added the "1" before dividing by "3"
okay from my answer.... not the internet.. 6+2 sort(21)/3.... where do i go from there?
well it should be a 6 on bottom
no i didn't add anything.
okay sure a 6. i thought it could be a 3 to narrow it down
so 6+or-2sqrt(21)/6
i dont see the problem, you posted answer in beginning... " i tried to use the quadratic formula and got sqrt(21)+3/3 "
so it is a 3 in the denominator!
\[\frac{6 \pm 2\sqrt{21}}{6} = \frac{2(3 \pm \sqrt{21})}{6} = \frac{3 \pm \sqrt{21}}{3}\]
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