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Consider the function $$y=x^2e^{-x^2}$$ The graph initially behaves as a parabola then in later part exponential part of it dominates; i.e., the graph looks exponential after maximum of the curve.

Actually this graph is related to Maxwell Boltzmann distribution graph. Please help me so that I can easily remember the property of this graph.

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  • $\begingroup$ Please confirm that the function resulting from my edit reflects the intent of your question. $\endgroup$ – abiessu Jun 24 at 2:59
  • $\begingroup$ Yes thanks a lot $\endgroup$ – shelton Benjamin Jun 24 at 3:02
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    $\begingroup$ MathJax hint: for multicharacter exponents, enclose them in braces, so e^{-x^2} gives $e^{-x^2}$. It works for many things, like subscripts and fractions as well. You can right click on any MathJax and choose Show Math As ->TeX commands to see how it was done. $\endgroup$ – Ross Millikan Jun 24 at 3:38
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You actually gave the mathematical explanation. The graph is below. Over the range $[-1,1]$ the exponential doesn't change that much-it is $1$ at the center and $\frac 1e \approx 0.3679$ at the ends. That is less than a factor $3$. The parabola is $0$ at the middle and $1$ at the ends, an infinite ratio. It dominates the product over this interval. As you get outside that interval, the exponential dominates. From $1$ to $3$ the parabola rises by a factor $9$, but the exponential drops by a factor $2980$, so it dominates.

enter image description here

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  • $\begingroup$ Thanks alot ross milikan you are such a nice guy $\endgroup$ – shelton Benjamin Jun 24 at 3:40
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$$f(x)=x^2e^{-x^2}\implies f'(x)=2x(1-x^2)e^{-x^2}, f''(x)=2x^2(x^4-5x^2+1)e^{-x^2}.$$ 1- Check that $f(x)$ is even and $f(0)=0, f(\pm \infty)=0,. f'(x)=0 \implies x=0, \pm 1, f''(\pm 1)<0, f'(x)$ does not change sign around $x=0$. So min at $x=0$ max at $x=\pm 1, f_{max}=e^{-1}, f_{min}=0.$ All these information helps plotting this function as ny @Ross Millikan in his answer here.

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  • $\begingroup$ Thank you sir please anyone can edit the question title and add Maxwell distribution relation so that it helps ones who search for it. I dont have edit privileges $\endgroup$ – shelton Benjamin Jun 24 at 4:00

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