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Find the maximum and minimum values of:

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the maximum and minimum values of

    f(x)=x^4e^-x over [0,10]

    2. Relevant equations

    3. The attempt at a solution

    From what I remember is you set the f' equal to 0. And that will give you two more points to test. and then you put the 0 and 10 number into the equation to find the values.

    When I put 0 or 10 in i get 0 for both.

    f'= (4x^3)(e^-x)+(x^4)(-e^-x)

    If i set that equal to 0; I really do not know how to find the what x is.
    Last edited: Aug 1, 2010
  2. jcsd
  3. Aug 1, 2010 #2

    Char. Limit

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    Gold Member

    Well, first, let's write this out...

    [tex]f'(x) = 4 x^3 e^{-x} - x^4 e^{-x} = 0[/tex]

    First, I'd recommend factoring.
  4. Aug 1, 2010 #3
    okay i see that now. so you get something like:


    so then


    and (4-x)=0

    so you get x=4

    and then i think the other one is ln0/-1 which is undefined if you set it to 0.
  5. Aug 1, 2010 #4
    Yep, okay i got it. Maximum is 4.6888 at x=4. and Min is 0 at x=0

    I don't know if (x3)(e^-x)=0 gives you ln0/-1 but either way 4 is the value I was looking for.

    Thank You
    Last edited: Aug 2, 2010
  6. Aug 1, 2010 #5

    Char. Limit

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    Hmm... I'm not sure where you're getting -log(0)... You know that if [itex]x^3 e^{-x} = 0[/itex], then either x^3=0 or e^(-x)=0, and only one of those has a solution.

    You've found a critical point at x=4, but there's one more.
  7. Aug 2, 2010 #6
    If you set x^3=0 You get x=0?
    If you set e^-x=0 You multiply both by (ln) so -x =ln(0) divide both by -1. That is what I thought? ln(0) / -1
  8. Aug 2, 2010 #7

    Char. Limit

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    Yes, but since [itex]ln(0)=-\infty[/tex], what you essentially said is that e^(-x)=0 for an x outside the standard real (and complex, incidentally) number system. So, it doesn't count as a root.

    Also, your domain was [0,10]. Negative infinity is outside that domain.
  9. Aug 2, 2010 #8


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    I would be inclined to write [itex]fsingle-quote(x) = 4 x^3 e^{-x} - x^4 e^{-x} = 0[/itex] as [itex]4x^3e^{-x}= x^4e^{-x}[/itex]. Since [itex]e^{-x}[/itex] is never 0, we can divide both sides of the equation by it and get [math]4x^3= x^4[/math]. If x is not 0, we can divide both sides by x to get x= 4.

    That tells you the two roots of the equation.
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