What is the equation of the line tangent to #f(x)= x^3e^x # at #x=2#?

1 Answer
Nov 23, 2016

#y-8e^2=20e^2(x-2)#

Explanation:

To find the equation of the tangent line, we need a point and a slope, then we can put it into point-slope form.

Point: #(2,f(2))#
#f(2)=(2)^3e^(2)=8e^2#
#(2,8e^2)#

Slope: #f'(2)#
#f(x)=(x^3)(e^x)#
#f'(x)=(x^3)(e^x)+(3x^2)(e^x)#
#f'(x)=(x^2)(e^x)(x+3)#

#f'(2)=((2)^2)(e^2)(2+3)#
#=(4)(e^2)(5)#
#=20e^2#

Point-slope form:#y-y_1=m(x-x_1)#
#y-8e^2=20e^2(x-2)#