What is the equation of the normal line of #f(x)=(1-x)e^(3x)-e^x# at #x=1#?
1 Answer
Nov 8, 2016
graph{((1-x)e^(3x)-e^x-y)(y+e-1/(e(e^2+1))(x-1))=0 [-3.844, 4.926, -3.28, 1.103]}
Explanation:
The general equation for normal is