# 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.