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

1 Answer
Oct 18, 2016

#y = 1 - x#

Explanation:

Let #y = f(x) = ln(x^2 - x + 1)/x#

#dy/dx = (x-2)/(x^2 (x^2-x+1))+(2-log(x^2-x+1))/x^2 #

Find the y coordinate at x = 1:

#y = ln(1^2 - 1 + 1)/1 = 0#

The slope, m, of the tangent line is #dy/dx# evaluated at #x = 1 #:

#m = (1-2)/(1^2 (1^2-x+1))+(2-log(1^2-1+1))/1^2 #

#m = 1#

The slope, n, of the normal line is:

#n = -1/m#

#n = -1/1#

#n = -1#

The normal line has a slope of -1 and passes through the point (1, 0). Using the point-slope form of the equation of a line, we obtain the following equation:

#y - 0 = -1(x - 1)#

#y = 1 - x#