anonymous
  • anonymous
The maximum value of the function f(x)=xe^(-x) is? Explain your answer please!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
take the derivative set it equal to zero solve for x
Astrophysics
  • Astrophysics
You will have to use the product rule \[(fg) = f'g+g'f\]
anonymous
  • anonymous
you get the derivative yet?

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anonymous
  • anonymous
or are you having problem setting equal to zero and solving?
anonymous
  • anonymous
so 1e^(-x)+xe^-x
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
that looks good factor out the common factor of \(e^{-x}\)
Astrophysics
  • Astrophysics
Ahh a wild misty appears!
anonymous
  • anonymous
I'm a bit lost and confused. haha, sorry.
misty1212
  • misty1212
lol \(\color\magenta\heartsuit\)
misty1212
  • misty1212
you got the derivative a bit wrong, you are off by a minus sign
misty1212
  • misty1212
the derivative of \(e^{-x}\) is \(-e^{-x}\) by the chain rule
Astrophysics
  • Astrophysics
You have the second part wrong should be \[e^{-x}-xe^{-x}\]
Astrophysics
  • Astrophysics
Oh this lag, misty is too fast
misty1212
  • misty1212
should have \[e^{-x}-xe^{-x}\] what @Astrophysics said
misty1212
  • misty1212
then you have a common factor in each term of \(e^{-x}\) factor it out and ignore it because \(e^{-x}\) is never zero
anonymous
  • anonymous
e^-x(1-x) I meant
misty1212
  • misty1212
lets go slow dear how would you factor \[x-yx\]?
misty1212
  • misty1212
or maybe \[\color\magenta \heartsuit-x\color\magenta\heartsuit\]
misty1212
  • misty1212
yeah that is right
misty1212
  • misty1212
and since \(e^{-x}\) is never zero, all you solve is \(1-x=0\) and you are done
anonymous
  • anonymous
-x(1-y)
anonymous
  • anonymous
Thanks very much. I understand this problem now. Sorry for being slow ^^
misty1212
  • misty1212
you are welcome \[\huge \color\magenta \heartsuit\]

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