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What is the slope of #f(x)=-xe^x/x^2# at #x=1#?

1 Answer

May 11, 2016

Slope of #f(x)# at #x = 1# is #0#

Explanation:

#f(x) = -x * e^x/x^2# #= -e^x/x# = #-e^x*x^-1#

#f'(x) = -(e^x*(-x^-2) + x^-1*e^x )# (Product, Exponential and Power Rules)

The slope of #f(x)# at #x=1# is given by #f'(1)#

#f'(1) = -(e^1 * (-1) + e^1 *1)#

#f'(1) = -(-e+e) = 0#

Hence: Slope of #f(x)# at #x = 1# is #0#

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