What is the slope of the line normal to the tangent line of #f(x) = xe^-x # at # x= 0 #?

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Answer 1 · 2017-01-13T15:00:38

#m_n = -1/(f'(x_0)) = -1#

The slope of the tangent to the graph of #y=f(x)# for #x=x_0# is #m_t = f'(x_0)#.

So the slope of the normal to the tangent is:

#m_n =-1/m_t = -1/(f'(x_0))#

In our case;

#f'(x) = e^(-x) -xe^(-x) = e^(-x) (1-x)#

For #x=0# the slope of the tangent is:

#m_n = -1/(f'(x_0)) = -1#

The equation of the normal line is:

#y= f(x_0) -1/(f'(x_0))(x-x_0)#

that is:

# y = -x#