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

Question

Ezekiel Alexander · Accepted Answer

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

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#

Ezekiel Alexander · Answer

undefined To find the equation of the normal line at x = -1, we need to determine the slope of the tangent line at that point and then find the negative reciprocal to obtain the slope of the normal line.

First, we find the derivative of f(x) using the product rule and chain rule:

f'(x) = [(-1/x^2)e^(-x^3+x^2)] + [(1/x)e^(-x^3+x^2)(-3x^2+2x)]

Next, we substitute x = -1 into f'(x) to find the slope of the tangent line:

f'(-1) = [(-1/(-1)^2)e^(-(-1)^3+(-1)^2)] + [(1/(-1))e^(-(-1)^3+(-1)^2)(-3(-1)^2+2(-1))]

Finally, we take the negative reciprocal of f'(-1) to obtain the slope of the normal line. Let's call it m:

m = -1/f'(-1)

Now we have the slope of the normal line. To find the equation of the line, we use the point-slope form:

y - y1 = m(x - x1)

Substituting x1 = -1 and y1 = f(-1) into the equation, we can simplify it to obtain the final equation of the normal line.