What is the equation of the normal line of #f(x)= x^3/(3x^2-2x)+6x# at #x = 2#?

The normal line is the line that is perpendicular to the line tangent to the function at a particular point. Imagine that you are skiing down a mountain. The mountain is your function, with its various hills and valleys. Your skis represent the slope of the mountain where you are; the skis represent the slope of the line tangent to your position along the mountain. You, while standing perpendicular to your skis, represent the slope of the normal line.

To find the slope of the normal line, we must first find the slope of the tangent line. For a function #f(x)# defined at #x = x_1#, the slope of the line tangent to #f(x)# at #x = x_1# is given by #f'(x_1)#. Ie, the derivative of a function evaluated at a particular point is the function's slope at that point.

We have

#f(x) = x^3 / (3x^2 - 2x) + 6x = x^2 / (3x - 2) + 6x#
#f'(x) = ((3x-2)(2x) - (x^2)(3)) / (3x-2)^2 + 6#
# = (6x^2 - 4x - 3x^2) / (3x-2)^2 + 6 = (3x^2 - 4x)/(3x-2)^2 + 6#

We could simplify further if desired, but at this point we can plug in our value of #x = 2#.

#f'(2) = (3(4) - 4(2)) / (6 - 2)^2 + 6#
# = 4/16 + 6 = 6 + 1/4 = 25/4#

Thus, the slope of #f(x)# at #x = 2# is #m = 25/4#. The slope of the normal line is the slope perpendicular to #m = 25/4#. This is given by the inverse reciprocal. So our normal line has a slope of #m = -4/25#.

To find our equation, we need more than just a slope, we also need a point. We are interested in the normal line passing through #f(x)# at #x=2#. Thus, our point is given by #(2, f(2)) = (2, 13)#.

With a point and a slope, we can generate the equation of our normal line using slope-intercept form.

#y = mx + b#
#y = -4/25 x + b#
#13 = -4/25(2) + b#
#13 = -8/25 + b#
#b = 13 + 8/25 = 333/25#

So our equation is #y = -4/25 x + 333/25#.