# Explanation for behaviour of graph of $y=x^2e^{-x^2}$ (Maxwell-Boltzmann distribution)

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.

• Please confirm that the function resulting from my edit reflects the intent of your question. – abiessu Jun 24 at 2:59
• Yes thanks a lot – shelton Benjamin Jun 24 at 3:02
• 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. – Ross Millikan Jun 24 at 3:38

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.
$$f(x)=x^2e^{-x^2}\implies f'(x)=2x(1-x^2)e^{-x^2}, f''(x)=2x^2(x^4-5x^2+1)e^{-x^2}.$$ 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.