What is the equation of the line normal to # f(x)=lnx-x^3# at # x=1#?

1 Answer
Jan 15, 2016

#2y = x -3#

Explanation:

First find the gradient of #f(x)# by differentiating. Then if the gradient is #m# the slope of the normal is #-1/m#
The rest of the equation can be found by substituting in the values at the given point.

#f(x) = lnx - x^3#
#f'(x) = 1/x -3x^2#
At # x=1#, #m = 1 - 3.1^2 = -2#

Therefore the slope of the normal is #1/2#

Going back to the original expression
f(1) = ln(1) - 1^3 = 0 -1 = -1#

The equation of the normal is #y=mx+c#
At the given point #-1 = 1/2(1) +c#
#:. c = -3/2#

The equation of the normal is
#y = 1/2x - 3/2#
or #2y = x -3#