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Homework Help: Function x - Finding the intervals, ranges, and points of inflection

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f be the function defined by f(x) = xe[tex]^{1-x}[/tex] for all real numbers x.

    a. Find each interval on which f is increasing.

    b. Find the range of F.

    c. Find the x-coordinate of each point of inflection of the graph of f

    2. Relevant equations

    I don't think there are any.

    3. The attempt at a solution

    It's been a while since I did calculus, so I might be a bit rusty on this. I'll take this 1 letter at a time.

    For part a, if f is incresaing, then its first derivative must be positive. Therefore, I took the derivative of f(x), using the product rule.

    f(x) = xe[tex]^{1-x}[/tex]
    f'(x) = x(1-x)(-1) + e[tex]^{1-x}[/tex]
    f'(x) = -x[tex]^{2}[/tex]+x-1+e[tex]^{1-x}[/tex]

    And ... this is where I get stuck. I'm not sure how to proceed from here for part a (I'll attempt the other 2 at a later date). :cry:
  2. jcsd
  3. Sep 23, 2009 #2
    The product rule is:
    (f*g)' = f'*g + f*g' where f and g are functions.
    Try finding f'(x) again.
  4. Sep 23, 2009 #3
    My notes say it's f*g' + g*f'. Or uv' + vu'.

    I've used this rule for other prod. rule problems and it seems to work all right.
  5. Sep 23, 2009 #4
    It doesn't matter what the order is; addition is commutative.

    Your error was when differentiating e1 - x.
  6. Sep 23, 2009 #5
    I thought e^(1-x) was -(1-x) because I bring down the (1-x) from the expnoent, then use the chain rule to get (-1) and just multiply the two together. Is it just (1-x)?
  7. Sep 23, 2009 #6
    You're missing a part when you differentiate e1-x. What is the derivative of ef(x) using the chain rule?
  8. Sep 23, 2009 #7
    Wait hang on ... is it ...

    f'(x) = -e[tex]^{1-x}[/tex]
  9. Sep 23, 2009 #8
    Now what is f'(x), all of it?
  10. Sep 23, 2009 #9
    Wow, I nearly ate my shorts. :biggrin:

    Anyway, here's the new f'(x).

    f ' (x) = e[tex]^{1-x}[/tex](1-x)

    I'm still not quite sure what to do next, can you sort of "poke" me in the general direction without actually giving away the answer since I'd like to learn this stuff.
  11. Sep 23, 2009 #10
    Now you want to find where f(x) changes from increasing to decreasing, or vice versa. What will the derivative be right where that change is?
  12. Sep 23, 2009 #11
    Just a quick question, but is finding what you said possible without a calculator, becuase my teacher says we can't use a calculator on this problem? Just wanted to make sure since don't you have to do some substitution of "x" to find where it's increasing/decreasing?
  13. Sep 23, 2009 #12
    Think of the graph of x2. What is the value of the derivative when it changes from positive to negative (the slope of the tangent line at the vertex)? No calculator needed.
  14. Sep 23, 2009 #13
    Oh ok .. good, x squared is easier for me. :)

    So the value of the derivative for x squared increases all the way through the graph (from left to right across the x axis). Derivative value at the origin is zero.

    The derivative graph is basically a straight line going northeast by going north 2 spaces and east 1 space.
  15. Sep 23, 2009 #14
    Now how do you find where the derivative of xe1-x changes from positive to negative, or vice versa?
  16. Sep 23, 2009 #15
    When x is more than than one, because 1-x means that the chain part of the derivative is negative, while the other part (e^(1-x)) is negative ... wait, nevermind that's 2 negatives. Hm ... I'm not sure.
  17. Sep 23, 2009 #16
    When is the derivative of xe1-x equal to 0? One one side of that point (or points) the derivative will be positive and the other side will be negative. Think back to x2.
  18. Sep 23, 2009 #17
    The derivative of xe[tex]^{1-x}[/tex] is zero when x=1.

    Ah, so the derivative of xe[tex]^{1-x}[/tex] is negative when x < 1 and positive when x > 1.
  19. Sep 23, 2009 #18
    Check the signs again for the derivative for x<1 and x>1.

    For b, how do you think you would go about finding the range? Is there a maximum or minimum value for the function?
  20. Sep 23, 2009 #19
    Ack! Sorry, it's actually positive when x < 1 (I then took out my graphing calculator to prove this, just for confirmation).

    For the range, I know the e graph doesn't have a negative y limit so the der. wouldn't have a POSITIVE y limit, right?

    (Gotta go offline now, but be back later, thanks for the help so far!)
  21. Sep 23, 2009 #20
    Think about this:
    Does f(x) go to positive infinity and negative infinity? If it doesn't go to one of them, say positive infinity for example, how do you find what the maximum value of f is? Think about the derivative...
  22. Sep 24, 2009 #21
    So, part b:

    The range of f is y =< 1.

    The thing is, I found this by my calculator, but apparently there's got to be a way to find this without calculator use.
  23. Sep 24, 2009 #22
    Wait, setting f(x) to equal zero or something should find the max value of f, I think.
    Last edited: Sep 24, 2009
  24. Sep 25, 2009 #23
    No, setting the derivative f'(x) equal to 0 will find a local max or min on the graph. Does y go to positive and/or negative infinity?
  25. Sep 26, 2009 #24
    Got this problem solved, thanks for your help, you got me on the right track. :)

    (I already handed in this problem for my hw. Yeah, I'm rusty with this stuff and forgot that whatever makes the first derivative zero is the max or min and whatever makes the SECOND derivative finds the points of inflection,etc.)
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