Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Urgend calculus question

  1. Feb 16, 2006 #1
    Hi

    I have been given the following assignment which has caused me some trouble:

    The function [tex]f(x) = x^b \cdot e^{-x}[/tex] where [tex]b \in \mathbb{R}_{+}[/tex]

    Determine if f has a minimum and a maximum, and find them.

    I know that the first step is determine f'(x) which is

    [tex]f'(x) = (\frac{b}{x} - ln(e)) \cdot e^{-x} \cdot x^b[/tex]

    Any hints what I do next ?

    Best Regards

    Mathboy20
     
    Last edited: Feb 16, 2006
  2. jcsd
  3. Feb 16, 2006 #2

    TD

    User Avatar
    Homework Helper

    If there is a minimum or a maximum, the derivative has to be zero there. Watch out though, a zero derivative doesn't necessarily imply an extremem (min or max) but it is a necessary condition. So find the values of x for which [itex]f'(x)=0[/itex].
     
  4. Feb 16, 2006 #3
    Do I then choose an abitrary b-value and then solve f'(x) = 0 ???

    Like lets say b = 1

    then [tex]f'(x) = -(ln(e) \cdot x -1) \cdot e^{-x}= 0 [/tex]

    Then [tex] x = \frac{1}{ln(e)} [/tex]

    If I then choose an abitratry b which lies in the inteval [tex] [1, \infty[ [/tex]

    I get that [tex]x = \frac{b}{ln(e)}[/tex]

    is that x then the maximum value for the given function?

    Best Regards
    Mathboy20

     
    Last edited: Feb 16, 2006
  5. Feb 16, 2006 #4

    TD

    User Avatar
    Homework Helper

    Rather then doing it for a particular value of b, do it in general (keep b as a parameter).
    You indeed get [itex]x = \frac{b}{\ln(e)}[/itex] but don't you think you could simplify [itex]\ln(e)[/itex] a bit? :wink:
     
  6. Feb 16, 2006 #5

    Yeah sure then x = b

    Can I then conclude that the parameter 'b' is the maximum value for the function ?

    Best Regards

    Mathboy20
     
  7. Feb 16, 2006 #6

    VietDao29

    User Avatar
    Homework Helper

    Be careful, when b is not 1, then there is one more value that makes f'(x) = 0. That value is x = 0. Do you know why?
    To check if it's the maximum or minimum value, one can try to take the second derivative of that function.
    If f''(x) > 0, and f'(x) = 0, then it's a minimum value.
    If f''(x) < 0, and f'(x) = 0, then it's a maximum value.
    Do you know this?
     
  8. Feb 16, 2006 #7

    TD

    User Avatar
    Homework Helper

    Although it's correct (this will be a maximum), you cannot conclude this from the zero derivative only. In order for that point to be a max (c.q. min), the derivative has to change sign arround that point (from + to - for a max and vice versa for a min).

    Alternatively, you could check the sign of the second derivative in that point. If that's negative, you've got a maximum while you'll have a minimum when that's positive.
     
  9. Feb 16, 2006 #8
    Hello again,

    Lets recap what I know.

    I'm given the function [tex]f(x) = x^{a} \cdot e^{-x}[/tex]

    where [tex]a \in \mathbb{R}_{+}[/tex] which means that [tex]a > 0[/tex]

    Futher I'm told that [tex]x \in [0, \infty[ [/tex]

    I required first to prove that there exists a minimum and a maximum value for f, and to determain those values.

    What would be the first logical step here?

    Sincerely Yours
    Mathboy20
     
    Last edited: Feb 16, 2006
  10. Feb 16, 2006 #9

    VietDao29

    User Avatar
    Homework Helper

    First, you should notice that:
    f(x) > 0 for all x > 0, and f(x) = 0 for x = 0, right? So what can you say about f(0)? (a maximum value or minimum value?)
    And the second hint is to follow TD's suggestion and prove that x = b is a maximum.
    Can you go from here? :)
     
  11. Feb 16, 2006 #10
    Dear TD,

    If I understand Your explaination correctly then I get the following

    first my f'(x) was wrong the right one is

    f'(x) = (b/x -1) * x^b * e^-x =0

    Then according to the definition then there exist a maximum if f''(x) < 0 and f'(x) = 0

    first condition:

    by choosing x = b

    then f'(b) = 0

    Second condition:

    f''(b) = -b ^ (b-1) * e^-b < 0 because b>0

    Then the maximum for f is x = b...

    There exist a minimum if f''(x) > 0 and f'(x) = 0

    The first point in the definition interval is x = 0, but this is unuserable since b/0 is not allowed.

    Then f doesn't have a mimimum???

    Sincerley Yours
    Mathboy20
     
    Last edited: Feb 16, 2006
  12. Feb 17, 2006 #11

    VietDao29

    User Avatar
    Homework Helper

    f'(x) = bxb - 1e-x - xbe-x
    -------------------
    f''(x) = b(b - 1)xb - 2e-x - bxb - 1e-x - bxb - 1e-x + xbe-x
    = e-x(b(b - 1)xb - 2 - 2bxb - 1 + bxb)
    f''(b) = be-b(b(b - 1)bb - 2 - 2bbb - 1 + bb) = e-b(-bb - bb - 1 + bb) = -bb - 1e-b < 0.
    Yes, so your work is correct.
    -------------------------
    However, you should also check the value of f(x) at the end point, too to see if it's a maximum or minimum there. As I told you before:
    f(0) = 0
    And f(x) > 0 for all x > 0
    So f(0) is a minimum value for f(x), since the domain of the function is [tex][0 ; \infty [[/tex]
    By the way, f'(x) = bxb - 1e-x - xbe-x = xb - 1e-x(b - x)
    So as long as b is not 1, f'(0+) = 0.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook