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

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Answer 1 · 2016-09-30T05:53:08

#2(1-e) ~= -3.43656#

#f(x)= -xe^x+2x#

#f'(x)= -xe^x+(-1)e^x+2# Product rule and power rule

#= -e^x(x+1)+2#

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

#f'(1) = -2e+2 = 2(1-e)#

#f'(1) ~= -3.43656#

This can be seen by the graph of #f(x)# in the region of #x=1# below:

graph{-xe^x+2x [-0.202, 2.498, -1.007, 0.343]}

Answer 2 · 2016-09-30T06:08:47

Slope at #x=1# is #-3.4365#

Given -

#y=-xe^x+2x#

Its slope is defined by its first derivative.

#dy/dx=[-xe^x(1)+e^x(-1)]+2#
#dy/dx=-xe^x-e^x+2#

Its slope at #x=1#

#dy/dx=-(1)e^1-e^1+2#
#dy/dx=-e^1-e^1+2=-3.4365#