What is the equation of the normal line of #f(x)= 1/xe^(-x^3+x^2) # at #x=-1#?

1 Answer
Yahia M.
Apr 6, 2018

#5e^2y+5e^4+x+1=0#

Explanation:

You start by finding the first derivative .
#y'=(-3x+2)e^(-x^3+x^2)#

by substituting with #x=-1# in both the function and the first derivative you get the y co_ordinate of the point that lies on the normal line and the slope of the tangent to the curve ( m )

#y=1/-1e^(2)#
so the y co_ordinate of the point =#-e^2#

and the slope of the tangent ( m ) at #x=-1# equals:
#y'=5e^2#
but We don't want the slope of the tangent, We want the slope of the normal.

The slope of the normal =#-1/m#

so it will be #-1/5e^-2#

and to get the equation of the straight line

#y-(-e^2)=-1/5e^-2(x-(-1)#

and by simplification you get:
#5e^2y+5e^4+x+1=0#

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