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Mathematics, 01.07.2020 16:01 FantasticFerret

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (Enter your answers as a comma-separated list of ordered pairs.) x = t^3 - 3t, y = t^2 - 4

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In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to...
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