What is the equation of the normal line of $f (x) = \frac{1}{1 - 2 e^{3 x}}$ $f(x)=1/(1-2e^(3x)$ at $x = 0$ $x=0$ ?

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Answer 1 · 2017-01-20T04:59:42

$x + 6 y + 6 = 0$ $x+6y+6=0$ . 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 $P (0, - 1)$ $P(0, -1)$ .

$f (1 - 2 e^{3 x}) - 1$ $f(1-2e^(3x))-1$ . 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.

What is the equation of the normal line of f(x)=1/(1-2e^(3x)f(x)=11−2e3xf(x)=1/(1-2e^(3x) at x=0x=0x=0?