What is the equation of the normal line of #f(x)=1/(1-2e^(3x)# at #x=0#?
1 Answer
Jan 20, 2017
Explanation:
graph{(y(1-2e^(3x))-1)(x+6y+6)(x^2+(y+1)^2-.04)=0 [-10, 10, -5, 5]}
The foot of the normal is ( as marked in the graph ) is
f'(1-2e^(3x))-6ye^(3x)=0#,
giving
So, the slope of the normal is #-1/f'=-1/6, and so, its equation is
x + 6y + 6 = 0.
See the graphical depiction.