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

1 Answer
Dec 18, 2017

x=1

Explanation:

By definition, the normal line must have a slope that is the opposite reciprocal of the slope of the tangent line at #x=1#.
Thus, to determine the slope of the normal line, we must first calculate the slope of the tangent line at x=1, which is just the derivative of #f# at #x=1# or #f'(1)#.

Step 1: Calculate #dy/dx#
#d/dx[(x-1)^2/(x^2+2)]#
#= (2(x^2+2)(x-1)-(x-1)^2(2x))/(x^2+2)^2#

Step 2: Find f'(1)
#f'(1)=(2(3)(0)-(0)^2(2))/(1^2+2)^3#
#f'(1)=0#

Step 3: Determine the equation of the normal line
Knowing that #f'(1)=0# tells us that there is the graph of #f# has a horizontal tangent line at #x=1#. Thus, the normal line must have a slope of #oo#, which is undefined. Because vertical lines are the only type of line with undefined slopes, #f# must have a vertical tangent line at #x=1#.

The equation for the vertical tangent line is #x=1#.