# Find an equation of the tangent line to the curve at the point (1, 1/e) ....(one, one over e)y=x^3e^-x ....y equals x to the 3rd power multiplied by e to the negative x power

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`f(x)= x^3 e^-x `

`==gt f'(x)= (x^3)(e^-x)' + (x^3)'(e^-x) `

`==gt f'(x)= -x^3e^-x + 3x^2 e^-x `

`==gt f'(x)= x^2 e^-x (3-x)`

==> Now we will find the slope of the tangent line at x=1 which is f'(1).

==> The slope `m= f'(1)= 1e^-1 (3-1)= 2/e`

==> Now we wil find the eqauation of the line given the point (1,1/e) and the slope 2/3

`==gt y-y1= m (x-x1)`

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

`==gt y= (2/e) x - 2/e + 1/e `

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

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