Web Analytics
Answered

Find an equation of the tangent line to the curve at the point (1,1/e). y=x^4e^-x

My attempt:

y'=(4x^3e^-x)-(x^4*e^-x)

y-y1=m(x-x1)

y-(1/e)=(4x^3)(e^-x)-((x^4)(e^-x))(x-1)

y=4x^4 e^-x - x^5 e^-x - 4x^3 e^-x + x^4 e^-x +1/e

Verified Answer :

Nirina7

we will use y-y1=m(x-x1), where m= f '(x)= 4x^3e^-x - x^4e^-x, f ' (1) =3e^-1
f  (1) =4e^-1
so the equation of tangent line tothe curve at the point (1,1/e) is 
y= 3e^-1(x-1) + 4e^-1, or y= 3e^-1(x)) + e^-1,

Related Questions