# What is the equation of the normal line of f(x)=1/(1-2e^(3x) at x=0?

Jan 20, 2017

$x + 6 y + 6 = 0$. See the normal-inclusive Socratic graph.

#### 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 $P \left(0 , - 1\right)$.

$f \left(1 - 2 {e}^{3 x}\right) - 1$. So,

f'(1-2e^(3x))-6ye^(3x)=0,

giving $f ' \times \left(- 1\right) - 6 \left(- 1\right) = 0 \to f ' = 6$, at $x = 0$.

So, the slope of the normal is -1/f'=-1/6, and so, its equation is

$y - \left(- 1\right) = - \frac{1}{6} \left(x - 0\right)$, giving

x + 6y + 6 = 0.

See the graphical depiction.