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

Question

Ezekiel Alexander · Accepted Answer

#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.

Christopher Agresta · Answer

undefined To find the equation of the normal line of f(x) = 1/(1-2e^(3x)) at x = 0, we need to determine the slope of the tangent line at x = 0 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) with respect to x:

f'(x) = (6e^(3x))/(1-2e^(3x))^2

Next, we evaluate f'(x) at x = 0 to find the slope of the tangent line:

f'(0) = (6e^(3*0))/(1-2e^(3*0))^2
      = 6/(1-2)^2
      = 6/1
      = 6

The slope of the tangent line at x = 0 is 6.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

m_normal = -1/6

Now, we have the slope of the normal line. To find the equation of the normal line, we use the point-slope form of a line and substitute the values of x = 0 and the slope:

y - y1 = m(x - x1)

Using the point (0, f(0)) on the curve, which is (0, 1/(1-2e^(3*0))), we substitute the values:

y - 1/(1-2e^(0)) = (-1/6)(x - 0)

Simplifying further:

y - 1 = (-1/6)x

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