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

Question

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

Elijah Abram · Accepted Answer

x=1

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#.

Ezekiel Alexander · Answer

undefined The equation of the line normal to f(x)=(x-1)^2/(x^2+2) at x=1 is y = -1.