What is the instantaneous rate of change of #f(x)=xe^x-(x-2)e^(3x-2) # at #x=-1 #?

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Answer 1 · 2016-01-17T06:29:13

#f' (-1)=8*e^-5#
#f' (-1)=0.053903575992684#

Find the first derivative #f' (x)# then evaluate #f' (-1)#
It goes like this;

#f(x) = x e^x-(x-2) e^(3x-2)#

#f' (x) =x e^x*1+e^x*1-[(x-2)*e^(3x-2)*3+e^(3x-2) *1]#
now use #x=-1#

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

#f' (-1)=0-(-3 e^-5 *3+e^-5)#

#f' (-1)=9e^-5-e^-5#

#f' (-1)=8 e^-5#

#f' (-1)= 0.053903575992669#