What are the critical values, if any, of #f(x) = xe^(2x)#?

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Answer 1 · 2017-04-13T02:35:53

Critical Value: #x=-1/2#

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

#f'(x)=1 cdot e^(2x)+x cdot 2e^(2x)=(1+2x)e^(2x)#

By setting #f'(x)=0#,

#(1+2x)e^(2x)=0#

By dividing both sides by #e^(2x)#,

#1+2x=0#

By subtracting 1 from both sides,

#2x=-1#

By dividing both sides by #2#,

#x=-1/2# (Critical Value)

I hope that this was clear.