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Graph e^-x^2

  1. May 15, 2014 #1


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    1. The problem statement, all variables and given/known data
    This may not be the correct place to ask but the problem gives a graph of f'(x)=e^-x^2 with a particular solution f(0)=0. It then asks you to find the limit of f(x) as it approaches infinity using a graphing calculator


    2. Relevant equations

    3. The attempt at a solution

    I would think I would have to integrate in order to solve this - but I have no idea how to integrate this. My Tnspire comes up with an error message. By fudging I am guessing the answer is .886, but that won't help me if I have to do this on an exam. It is obvious from the graph that it is 1 or less. If someone can point me to some instructions or something I would be grateful.
  2. jcsd
  3. May 15, 2014 #2


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    ##erf(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt##
  4. May 15, 2014 #3


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    Okay - I get the right side, but have never seen the er f(x). Is this problem overkill for high school Calc AB or have I really missed something. If I plug that is I get .999991, so would that make the limit 1?
  5. May 15, 2014 #4


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    That's a good estimate. The exact value is ##\sqrt{\pi}/2##. However, it's not at all trivial to calculate analytically. Here is one proof:


    Since this is posed as a graphing calculator problem, I'm guessing that 0.886 is a perfectly reasonable answer.

    By the way, your calculator will probably like the expression better if you introduce some parentheses: e^-(x^2)
  6. May 15, 2014 #5


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    Thanks - I was thinking I needed to integrate the equation in order to determine the limit. I did graph f '(x) = e^-(x^2). If I graph that in the area for slope fields, I get similar slope field, but when I run through the tables it appears y is approaching .93 and that is not one of the options. I think I am missing something. Exactly how am I supposed to use a slope field to determine a limit without knowing the original equation?
  7. May 15, 2014 #6


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    Graphing erf

    Well some points of interest are known from the derivatives:

    Since [itex]f^{\prime}(x)=e^{-x^2}\in\left( 0,1\right]\forall x[/itex] and, more to the point, [itex]f^{\prime}(x)\neq 0[/itex] so [itex]f(x)[/itex] is always increasing and has no critical points. Furthermore, [itex]\lim_{x\rightarrow\pm\infty}f^{\prime}(x)=\lim_{x\rightarrow\pm\infty}e^{-x^2}= 0[/itex] so the curve [itex]f(x)[/itex] has tangents whose slopes approach 0 as [itex]x\rightarrow\pm\infty[/itex], so, if [itex]x=\pm\infty[/itex] were points, they would be the global maximum (at [itex]x=+\infty[/itex]) and global minimum (at [itex]x=-\infty[/itex]), respectively.

    Now on to [itex]f^{\prime\prime}(x)=-2xe^{-x^2}[/itex]: setting equal to 0 gives x=0 as the only inflection point (a point where f(x) changes concavity) and f''(x) has the opposite sign as x, so f'' indicates that f(x) is "concave up" (think how [itex]y=x^2[/itex] is bowed) for [itex]x\in \left( -\infty , 0 \right) [/itex] and that f(x) is "concave down" (think how [itex]y=-x^2[/itex] is bowed) for [itex]x\in\left( 0,+\infty \right) [/itex].

    It should be obvious that f(0)=0, and we also so have the horizontal asymptotes [itex]y=\lim_{x\rightarrow\pm\infty}f(x)=\pm\frac{\sqrt{\pi}}{2}[/itex] (I stole this from a previous post by jbunniii, and he's right about the proof).

    Put all this together and sketch f(x). plot note that this plot is normalized by the scale factor of [itex]\frac{2}{\sqrt{\pi}}[/itex].
    Last edited: May 15, 2014
  8. May 15, 2014 #7


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    Thanks to all who responded. The question I had was how to get the 2√pi as the horizontal asymtotes from the data given. I now realize that erf(x) is not just an error but an actual function. Really hate it when they quiz us over "stretch stuff" that we had no real way of knowing. Extra thanks to jbunniii for verifying my "fudge guestimate" was a good answer. Still may be some function on my graphing calculator to give me the limit that I haven't figured out, but at least I understand it.
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