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Sketching a curve to how unique solutions

  1. Jul 7, 2012 #1
    Sketch the graph of the equation x - e^(1-x) - y^3 = 0, Show that for each x there is a unique y satisfying the equation.

    Attempt:

    So the first thing I did was isolate y in order to put the equation in a form to graph (somewhat). did that and got y = (x-e^(1-x))^1/3.

    Got the y-int: (0, -e^1/3), x-int: (1,0) [I didn't get this, this was in soln, how'd they get?]

    Now after my attempt to graph it, I look at the solution and they say:

    we see that the graph is asymptotic to the curve y= x^1/3 as x-->∞ and asymptotic to the curve y = -e^[(1-3)/3] as x --> -∞

    How is there any asymptotic behavior? The expression doesn't indicate any restrictions. This is based on Implicit Function Thm by the way
     
  2. jcsd
  3. Jul 7, 2012 #2

    Mentallic

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    They got it by observation. If you can understand the shape of the graphs y=x and y=-e1-x, then adding those together (because we are trying to solve [itex]x-e^{1-x}=0[/itex]) must give us an x-intercept at a reasonably small value of x that is greater than 0. x=1 is what we're looking for.


    Well what happens as [itex]x\to\infty[/itex] for [itex]e^{1-x}[/itex]? And try comparing ex with x as [itex]x\to\infty[/itex], which value grows faster?
     
  4. Jul 7, 2012 #3




    Well for the first part as [itex]x\to\infty[/itex] for [itex]e^{1-x}[/itex] the function will tend to 0, but if I'm studying the behavior as x-->∞ don't I have to take into account the other portion of the expression i.e: x from [x-e^(1-x)]? so the e^(1-x) will go to 0, but that other x goes to ∞. That's why I'm confused I know e^x will go to ∞ faster than x alone but that would force my expression to "break" the asymptote.
     
  5. Jul 7, 2012 #4

    Mentallic

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    Yes exactly, but the whole point of an asymptote is that one function tends to get closer and closer to another.
    In this case, as [itex]x\to\infty[/itex], [itex]e^{1-x}\to 0[/itex] therefore [itex]x-e^{1-x}\to x[/itex], which is why your function is asymptotic to [itex]y_1=x^{1/3}[/itex] as [itex]x\to \infty[/itex].

    I understand what you're thinking, and in fact you're right. Asymptotes are defined as being curves that approach each other. They need to get arbitrarily close to a distance of zero between them.
    I believe the point your teacher was trying to make is that if you're looking to sketch that graph, then you can get an idea of what happens as [itex]x\to -\infty[/itex] because it'll look as though the graph is asymptotic to [tex]y_2=e^{\frac{1-x}{3}}[/tex] (but it's not!) because the fractional difference between the values of [tex]y=\left(x-e^{1-x}\right)^{1/3}[/tex] and y2 will become very small for x values <0 (but the difference won't approach zero).
     
  6. Jul 7, 2012 #5
    Ok, in this instance I see what your talking about, now in general from how you went about explaining things I gather what may arise sometimes is that I have to refer to the graphs of functions I'm familiar with i.e e^x, x^1/3, etc and relate them to my given function to give me an idea of the behavior of the "complex" function to graph? It's really appearing more and more that we have to use previously proven facts in order to "model" these complex looking functions. Am I on the right train of thought?

    Thanks by the way.
     
  7. Jul 7, 2012 #6

    Mentallic

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    Yep, pretty much. You need to have a firm grasp with the basic functions, such as your polynomials, exponentials, logarithms, trig etc. and then as was shown in this problem, it might not be clearly obvious how you'll be needing to graph such functions, but what you can do is consider intercepts, asymptotes, turning points and anything else you feel is deemed necessary. You may even want to plug in some values to get a better idea of what's going on.

    Use all the tools you can.
     
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