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

Question

Scarlett Allison · Accepted Answer

#y=1/(3e^9)x-1/e^9+e^9# If tangent line is #y=kx+n# then #k=f'(x_0)# where #x_0=3#. #f'(x)=e^(x^2/(x-2)) * (2x(x-2)-x^2)/(x-2)^2# #k = f'(3)=-3e^9# Equation of the normal is #y-y_0 = k_1(x-x_0)# where: #x_0=3 => y_0=f(x_0)=f(3)=e^9# and #k*k_1=-1 => k_1=-1/k = 1/(3e^9)# #y-e^9=1/(3e^9)(x-3)# #y=1/(3e^9)(x-3)+e^9# #y=1/(3e^9)x-1/e^9+e^9#

William Abdalla · Answer

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

To find the slope of the tangent line at x = 3, we can use the derivative of f(x). Taking the derivative of f(x) with respect to x, we get:

f'(x) = (2x(x-2) - x^2)/(x-2)^2 * e^(x^2/(x-2))

Evaluating f'(x) at x = 3, we have:

f'(3) = (2(3)(3-2) - 3^2)/(3-2)^2 * e^(3^2/(3-2))
       = (6(1) - 9)/(1)^2 * e^(9/1)
       = (6 - 9)/(1) * e^9
       = -3 * e^9

Therefore, the slope of the tangent line at x = 3 is -3 * e^9.

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

m_normal = -1 / (-3 * e^9)
         = 1 / (3 * e^9)

Now that we have the slope of the normal line, we can use the point-slope form of a line to find the equation. Since the normal line passes through the point (3, f(3)), we substitute these values into the equation:

y - f(3) = (1 / (3 * e^9))(x - 3)

Simplifying further, we have:

y - f(3) = (x - 3) / (3 * e^9)

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