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Find an equation of the tangent line to the curve at the point (1, 1/e).

y = (x^3)(e^−x)

2 answers

y = (x^3)(e^−x)
y' = -x^2 e^-x (x-3)
so, y'(1) = 2/e

Now you have a point and a slope, so the line is

y - 1/e = 2/e (x-1)

see

http://www.wolframalpha.com/input/?i=plot+y+%3D+%28x^3%29%28e^%E2%88%92x%29%2C+y%3D2%2Fe+%28x-1%29%2B1%2Fe+for+0.5+%3C%3Dx+%3C%3D+1.5
I took my derivative wrong when I did this on my own. Thank you!
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