What are the critical values, if any, of #f(x)=xe^(-3x)#?

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Answer 1 · 2015-11-22T18:28:59

#P(1/3, 1/(3e))#

First we derivate this, so

#y = xe^(-3x)#

#ln(y) = ln(x) -3x#

#1/y*dy/dx = 1/x - 3#

#dy/dx = e^(-3x) -3xe^(-3x)#

Critical points are those where the derivative is 0, so

#0 = e^(-3x) - 3xe^(-3x)#

Or

#3xe^(-3x) = e^(-3x)#

Because the exponential is never 0 you can divide it out

#3x = 1#
#x = 1/3#

The function has a critical point, in #P(1/3, 1/(3e))#