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Given the function, show that the derivative

  1. Oct 21, 2013 #1
    Hi, I have the following assignment:

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

    Given the function f(x) = x^2*e^-x, x>-1

    a)

    Show that f'(x) = 2xe^-x - x^2 * e^-x

    b)

    we're going to examine the properties of the function f.
    1: find the limits when x -> ∞
    2: find potential global and local max/min.
    3: draw the graph (this is pretty self explanatory and I'll most likely be able to do this without any problems, I'll just add it to get the complete exercise.)

    c)

    The line l tangents the graph to the function f at point given at x = 1.
    Find the equation to the line and show that the line crosses the graph to x at origin.

    d)

    Find the area (flat(?)) defined by the graph to f(x) and the line y = e^-1 * x by calculations.


    2. Relevant equations

    I'm currently at part a) where I'm struggling with the - sign in front of x^2 * e^-x.

    3. The attempt at a solution

    f(x) = x^2 e^-x = u*v = u'v+uv'
    u = x^2 u' = 2x
    v = e^-x v = e^-x

    f'(x) = 2x*e^-x + x^2 * e^-x

    Why am I getting + instead of -? Does it have anything to do with the x > -1?

    Thanks for any input.
     
  2. jcsd
  3. Oct 21, 2013 #2

    FeDeX_LaTeX

    User Avatar
    Gold Member

    ##\frac{d}{dx}(e^{-x}) \neq e^{-x}##.
     
  4. Oct 21, 2013 #3
    Thanks!

    e^-x = e^-x * (-1). I thought any derivative of e was the same as the original expression/function. I guess this makes a lot more sense as the derivative of e^x is e^x because x = 1. Lesson learned.

    Moving on to my next problem (might post here again tomorrow if I'm stuck).
     
  5. Oct 21, 2013 #4

    Mark44

    Staff: Mentor

    No, this isn't true. d/dx(ex) = ex, but for different exponents you need to use the chain rule.
    It has nothing to do with the value of x, but it does have to do with the coefficient of x. d/dx(ekx) = k * ekx
     
  6. Oct 24, 2013 #5
    I found the limit by doing so:

    lim x->∞ x^2 * e^-x = x^2 * 1/e^x = x^2/e^x = 0. That should be self explanatory since e^x will increase a lot steeper than x^2. (mind you, I have lim x->∞ in every step).

    Now I'm stuck at finding global/local critical points (maxima/minima). I do this by identifying critical points, and that can be done by finding the derivative of the function. This was done in exercise 1a.

    Critical points are the point at which the derivative of the function is equal to 0 or undefined, as far as I know.

    As my function is defined for x > -1, I reckon this is useful to find one critical point. The other one is found when the derivative is equal to zero. I'm not sure whether or not there are more than one critical point for both maximum and minimum values, but I'm going to assume so as the assignment clearly states maxima and minima.

    Anyhow, this is what I've done so far:

    f'(x) = 2xe^(-x) - x^(2)e^(-x)

    f'(x) = 0 -> 2xe^(-x) - x^(2)e^(-x) = 0

    I rewrite this to get the following:

    xe^(-x) * (2-x) = 0.

    Now I'm completely stuck! Mind you, I'm well aware that I might've made mistakes before this point.

    Something tells me that for this function to be 0, either xe^(-x) has to be zero, or (2-x) has to be. But I still can't figure this out. I think I'm experiencing what some Americans call a brainfart.

    Any suggestions to help me out, both with the one I've attempted to solve myself, and the other critical point(s) that I have no idea how to begin solving? Thanks in advance.
     
  7. Oct 25, 2013 #6
    Yes, if you multiply a bunch of things together and get 0 one of the factors has to be 0. You have 3 factors, not 2, and ##e^{-x}## is never 0. The other two factors can be 0.
     
  8. Oct 26, 2013 #7
    Thanks a lot for your reply.

    I'm confused about the term factors then. The way I think of it, the only factors I have is xe^(-x) and (2-x).

    Anyhow, xe^(-x)(2-x)=0 -> x=0.

    So I have a critical point at x=0?

    I do not have my 'proper' calculator here, that I use to draw graphs with. Is that the only way I can determine whether or not this function have its critical point at local/global minimum or maximum? If it is, I'm sure Google can help me find a site that allows me to graph a function.

    I must admit, I'm slightly confused about this particular problem. Any help is greatly appreciated.
     
  9. Oct 26, 2013 #8
    Yes, you do have a critical point at x = 0 (and x = 2). This comes directly out of your equation ##xe^{-x}##(2-x) = 0. You have 3 factors, not 2: x, ##e^{-x}## and (2-x). It's as if I wrote 3x(x-2) = 0. You know in that case that x = 0 or x = 2. The 3 is never going to be zero, so you ignore it for this purpose.

    What you perhaps don't know is that ##e^{-x}## will never be 0, just as 3 will never be 0.

    Re whether each critical point is a local min or max, you get that from looking at the 2nd derivative where you have one (which you do in this case). Check your book for the 2nd derivative test.

    As to whether a local min or max is a global min or max, that depends, and you have to work it through, but you are not up to that yet.

    Graphing never hurts, but it is not essential for this problem.
     
  10. Oct 27, 2013 #9
    Thanks a lot for your reply again.

    So I have two critical points, x= 0 and x= 2.

    When I take the 2nd derivative, I get f''(x) = -2e^(-x)-x^(2)e^(-x)

    I then replace x with 0 and 2 and get: f''(0)=-2 and f''(2)=-0,81. That means that both of these critical points are local minimum points? As my function goes to x>-1, that means that only f''(2) is correct?

    Will my answer to this problem be that the only global or local minima or maxima is the local minimum point x=2?

    Thanks for any input. Even though I don't see the use for this in real life, at least not yet, this is incredibly interesting to learn about!
     
  11. Oct 27, 2013 #10
    The first problem is that I don't think you took the 2nd derivative correctly. That may be due to thinking you have 2 factors when in fact you have 3. x##e^{-x}## is 2 factors, and you still have (2-x) as the 3rd.

    Consider this formula: d/dx[f(x)g(x)h(x)] = f'(x)g(x)h(x) + g'(x)f(x)h(x) + h'(x)f(x)g(x). This is an extension of the product rule to 3 functions. That is what you have to use to take the second derivative. Then if you get your algebra right, you will have the correct second derivative.

    You are right that you want to look at f''(0) and f''(2). But the theory says if a is a critical point and f''(a) >0 you have a min; f''(a)<0 a max; f''(a) = 0 you have neither. To remember this I use the function g(x) = ##x^2##. It has a minimum at x = 0 and g''(0) = 2.

    Try again with the second derivative, and when you have that much settled down we'll tackle the global/local issue.
     
  12. Oct 27, 2013 #11
    Again, thanks a lot for your reply!

    Hopefully I've understood you correctly, although I haven't seen this method to solve for 2nd derivatives before. This is what I've done:

    f''(x) = d/dx[f(x)g(x)h(x)] = f'(x)g(x)h(x) + g'(x)f(x)h(x) + h'(x)f(x)g(x)

    where:

    f(x) = x -> f'(x) = 1
    g(x) = e^(-x) -> g'(x) = -e^(-x)
    h(x) = (2-x) -> h'(x) = -1

    f''(x) = 1*e^(-x)*(2-x) + (-e^(-x))*x(2-x) + (-1)e^(-x)*x
    = e^(-x)*(2-x) - xe^(-x) * (3-x)

    Using this 2nd derivative I get the following:

    f''(0) = 2 > 0 -> minimum point
    f''(2) = -0,27 < 0 -> maximum point
     
  13. Oct 27, 2013 #12
    Yes, you got the f'' right. And you got the f''(0) right. For f''(2) I got -2##e^{-2}## which is certainly negative -- apparently you worked it out. That is fine, but wasn't necessary because we only wanted to know if it is negative, not exactly what it is.

    Moving on to the question of global vs local extrema, you need to look at the original function f(x) = ##x^2e^{-x}##. At x = 0 you have a minimum. What is its value? Is there any other real number a for which f(a) < f(0)? If there is, your minimum is local. If not, your minimum is global. To deal with the maximum at 2, you ask the same kinds of questions.

    In both cases, be sure to check the value x = -1. The reason is that max and min can occur on boundaries as well as critical points.

    Finally, can you use Latex -- it is much easier to read. At least enclose things like x^2 in ## (two on each side) so they will render nicely.
     
  14. Oct 27, 2013 #13
    Alright, I'll keep that in mind.
    For f(0) I have the value 0, and at f(2) I have the value [itex]\frac{4}{e^2}[/itex] = 0,54.

    By f(a) you mean any value for x? Unfortunately I can't remember how to solve this mathematically, and my book doesn't explain it properly in my opinion, but I'll manage to figure it out.
    When I check the value x = -1, I get e (2,718..). Does this mean that my critical points, my maximum and minimum points, are within this function ( x > -1)?
    Of course! I completely forgot how to use it, and then forgot to ask about it, but when you said Latex I remembered how I did it.

    I really appreciate your help.
     
  15. Oct 27, 2013 #14
    Yes, when I said f(a) I meant that a could be any value of x.

    When trying to figure out whether you have a local max or min, you have to look at the function as a whole.

    For example, what does the function do when x < 0? Does it go up? go down? bounce around? Check it out. Then look to the right of 0. Does the function go negative anywhere?

    With respect to the point at x = 2, you have a value for it and again the question is what is the function doing as you move away from x = 2. Check out what it is doing for x > 2. Is it going up? down? what?. Then, of course look at the situation where x < 2. Of course you know f(0) = 0 and f(2) > 0, so it has to get from ##4/e^2## down to 0 somehow. Does it just slide down, or does it first go up and then come down?

    Re the value at -1, you have to check it because you are trying to find out whether 0 and 2 are global or local min and max. If f(-1) < f(0) then f(0) is a local min but not a global min. And the other way for f(2).

    Again, max and min can occur on a boundary as well as at critical points.

    Can you finish up now? (Respond if there are more questions.)

    (By the way, around here we use a . for decimal point instead of a , -- that is 3.5 not 3,5. I was a bit confused for awhile about what you wrote. If your readers are mostly American, you might want to switch to a . for your posts).
     
  16. Oct 28, 2013 #15
    Thanks once again.

    I thought I didn't have any more questions, but I guess it's better to be safe than sorry.

    I now have the following: f(0) = 0. Since no value a can make the following statement true: f(a) < f(0), then f(0) is a global minimum.

    f(2) = [itex]\frac{4}{e^2}[/itex]. Since f(-1) > f(2), f(2) is a local maximum.

    Does this mean that f(-1) is the global maximum? Thus, all global/local max/min points for this function is f(-1), f(0 and f(2)? Or do I just have the two? This is where I think you mean that max/min points can occur at boundaries, meaning I have three points (local min, local and global max).

    I now have c) and d) left.

    I will however have a go at c) before I ask for help. If I need help, should I still post in this thread yes?

    Again, thank you so much for your help. Highly appreciated.
     
  17. Oct 28, 2013 #16
    Your reasoning is all correct. Yes f(-1) is the global maximum for this problem. When you write it up, make sure you explain that x = 2 cannot be a global max on this interval because of the boundary at x = -1.

    Notice that if the -1 boundary had not been there, f(x) would have climbed to ##\infty## on the the negative side.

    I think you can easily sketch the graph from this knowledge.

    Certainly, post again if you have further questions.
     
  18. Oct 28, 2013 #17
    The feeling of mastering a math problem that appeared impossible at first is great. Just great. And not only have I done it, I understand it, and I can explain it rather detailed in my opinion. I'm happy.

    Alright, I've done c, and this is my result:

    f(x) = ##x^2e^{-x}##
    f(1) = ##e^{-1}## = [itex]\frac{1}{e}[/itex] = 0.36

    f'(x) = ##xe^{-1}##(2-x)
    f'(1) = [itex]\frac{1}{e}[/itex]

    y = [itex]\frac{1}{e}[/itex]x + b

    The point I am finding the tangent line l for: (1,[itex]\frac{1}{e}[/itex])

    [itex]\frac{1}{e}[/itex] = [itex]\frac{1}{e}[/itex]*1 + b -> b = [itex]\frac{1}{e}[/itex] - [itex]\frac{1}{e}[/itex] -> b = 0.

    y = [itex]\frac{1}{e}[/itex]x -> y = 0 when x = 0. <- shows that the tangent line l crosses f(x) at origin.

    Using y = [itex]\frac{1}{e}[/itex]x I can also easily draw a graph.

    This seems perfectly fine to me.

    However, problem d is more difficult. Find the area (flat(?)) defined by the graph to f(x) and the line y = e^-1 * x by calculations. This is obviously translated, but I suspect you understand it better if I say the surface area.

    Any idea how I can go about to solve this? I've tried using the internet, but I haven't come across anything helpful yet.

    Edit: I just realized, surface area is not what I'm looking for as far as I'm aware. I'll try figure out a more accurate description.

    Edit2: A more correct description would be the following, I hope: find the area of the space enclosed by the function f(x) = ##x^2e^{-x}## and the tangent line y = [itex]\frac{1}{e}[/itex]x.

    My own attempt to follow below.
     
    Last edited: Oct 28, 2013
  19. Oct 28, 2013 #18
    It's nice to hear you are happy. I too find a great satisfaction in mastering something that started out difficult.

    You computed the line correctly in part c. You can save a little time doing these if you consider this: the equation of the line through point (a,b) with slope m is y - b = m(x-a). Maybe try it out for this case, and see it is the same.

    So for part d, you are looking for an area. A surface area would be on a surface, such as a ball. This is a plain area.

    You have sketched the graph of f(x). Now on that same graph, sketch y = x/e. What you will see is that the graphs cross at x = 0, which you already knew. They also cross at some x = a > 0 (and you know what a is too).

    So between the line y = x/e and curve y = x##^2e^{-x}## -- starting at 0 and running to a -- there is a little gap -- that is the area you are trying to compute. Can you compute the area under y = x/e on this interval? Can you compute the area under y = x##^2e^{-x}## on this interval? If so, then you only need to subtract them, and you will have the area of your gap.

    This is not so difficult because both functions are above the x-axis. It is just a little sneakier if part is above and part is below, but you are not up to that yet.
     
  20. Oct 28, 2013 #19
    In c, I found out that the tangent line l crosses f(x) at 0 and 1.

    f(x) = ##x^2e^{-x}##
    g(x) = [itex]\frac{x}{e}[/itex]

    0[itex]\int[/itex]1 (##x^2e^{-x}## - [itex]\frac{x}{e}[/itex]) dx = [##x^2e^{-x}## + ##3e^{-x}## - 2x - [itex]\frac{x^2}{2e}][/itex][itex]\stackrel{1}{0}[/itex]

    = ##4e^{-1}## - [itex]\frac{e^{-1}}{2}[/itex] -2 = -0,71.

    I was half way through this solution before I saw that you posted, but I decided to finish off before reading. I'm not sure if that was a good idea.

    Edit:

    As far as I understand, I'm on the right track by taking the integrals of the functions, and subtracting 0 from 1 - as I've attempted to do above. I doubt I've done it correctly though.

    Am I on the right track, or did you mean something else by computing the area?
     
    Last edited: Oct 28, 2013
  21. Oct 28, 2013 #20
    You are doing the right kind of thing. I didn't know how much you knew. The first thing is the line x/e is above the curve, so you want to put that first and subtract f(x), not vice versa. That's why you came out with a negative area; we usually consider area to be positive.

    Next, I don't think you found the right anti-derivative for x##^2e^{-x}##. If you differentiate what you've put down there you will see.

    It never hurts to work on problems by yourself. Even if you are wrong, you will learn something.

    Best wishes ...
     
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