Let f(t)=4 t^2 + 4/t^2. Find the slope of the curve at t = -3.

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Let f(t)=4 t^2 + 4/t^2. Find the slope of the curve at t = -3.

Mathematics
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f(t)'=8t -(8)/t^3 for t=-3 -24-8/27 -640/27=-20
Ooop it is not -20 -640/27=-23.7037
yeah I got df/dt=8t-2/t^2 so t=-3 => 8*(-3)-2/(-3)^2 -24-2/9 ! but i dont trust my work

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the second term when you differentiated;\[4 \over t ^{2}\]is\[4 t ^{-2}\]becomes \[-8 t ^{-3}\]
or\[-8\over t ^{3}\]
but I would still have to plug in -3 for t
Yes indeed.
wait was I on the right track to getting the answer?
that would be -8/27 for that term and 24 for the first term I see what you mean it can't be negative with 24 being positive!! lol, I better take a nap.
but -640/27=-23.7037 is the answer
right?
Oh yes, thats right as t=-3 so it would be -24 and a negative 8/27 that means we have to combine the -24 and the -8/27 giving -648/27 - 8/27=-656/27= -24.29
Looking at your next to last post. How did you get -640/27? Wouldn't -24 X 27 =-648 ??
It would appear so, if I have kept track of the signs.
But maybe I messed up let me think about this?
lol take your time
I think I did mess up on that 2nd term when I plugged in the minus 3\[-8\over t ^{3}\]\[-8\over (-3)^{3}\]\[-8\over -27\] or POSITIVE 8/27 lol there it is.
So we ADD the 8/27 to the negative 648/27 giving -640/27 So sorry paunic88, I did stray off the path of keeping track of signs.
ok so the final anwser is -640/27 right? (ill compute it later)
-3 X -3 X -3 = -27 lol
Yes.
Good luck with your studies,and don't err with signs lol
ok thank you very much for the help
ur welcome.

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