Is #f(x)=xe^x-3x # increasing or decreasing at #x=-3 #?

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Answer 1 · 2016-01-04T17:21:21

The derivative at #x=-3# is negative, so it's decreasing.

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

#f'(x)=(x*e^x-3x)'=(x*e^x)'-(3x)'=#

#=(x)'e^x+x*(e^x)'-(3x)'=1*e^x+x*e^x-3=#

#=e^x*(1+x)-3#

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

At #x=-3#

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

Since #2/e^3+3# is positive, the minus sign makes:

#f'(-3)<0#

The function is decreasing. You can also see this in the graph.

graph{x*e^x-3x [-4.576, -0.732, 7.793, 9.715]}