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

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

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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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