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OpenStudy (ribhu):
=e^(lnx^2)
OpenStudy (ribhu):
f'x = 2x
OpenStudy (anonymous):
I don't understand??
OpenStudy (ribhu):
prop of log
OpenStudy (anonymous):
so how do you get f'(3)?
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OpenStudy (ribhu):
put x=3 in f'x =2x
OpenStudy (anonymous):
so would the answer be 6:)
OpenStudy (anonymous):
???
OpenStudy (ribhu):
yups i have shown each and every step\[a^{\log_{a}x } = x\]
OpenStudy (anonymous):
Thanks:)
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OpenStudy (ribhu):
this is property of log whic's being applied
OpenStudy (anonymous):
The position of a particle moving along a straight line at any time t is given by s(t)=t^3+9t^2-27. What is the velocity of the particle at the time when the acceleration is zero?
OpenStudy (ribhu):
acceleration = s"(t)=0, get a value of t then put in s'(t) which would be velocity
OpenStudy (anonymous):
so would s"(t)=6t+18 ?
OpenStudy (anonymous):
......then set is equal to 0 to get -3??
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OpenStudy (ribhu):
yaa there seems something wrong in the question
OpenStudy (anonymous):
why?? -3 is an answer choice?
OpenStudy (ribhu):
time can not be negative
OpenStudy (anonymous):
then how would we get the correct answer?? b/c I'm pretty sure -3 is right......idk though
OpenStudy (ribhu):
-3 comes from s"t = 0
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OpenStudy (anonymous):
yes...
OpenStudy (ribhu):
and u've to calculates't at that t at which s"t -0
OpenStudy (anonymous):
I set s''t=0......the answer was -3
OpenStudy (ribhu):
yaa so t=negative not possible
OpenStudy (anonymous):
okay....so do you know what we do from here?
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OpenStudy (ribhu):
the quest is probably having a mistake.
OpenStudy (anonymous):
I read over the ? & everything seems right?...do you want to move to the nxt question? I'll just bubble -3 as my answer...:)
OpenStudy (ribhu):
its asking velocity which should be zero at t=-3, velo = -27 from this data
OpenStudy (anonymous):
yes....-27
OpenStudy (anonymous):
?
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