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

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Answer 1 · 2017-01-20T04:59:42

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

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(0, -1)#.

#f(1-2e^(3x))-1#. So,

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

giving #f' xx (-1)-6(-1)=0 to f' = 6#, at #x = 0#.

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

#y-(-1)=-1/6(x-0)#, giving

x + 6y + 6 = 0.

See the graphical depiction.