Infimum and supremum of function

Question

Let $a >0$ and consider function $g : \mathbb{R} \rightarrow \mathbb{R}, g(x) = x e^{-ax^2}$. Find infimum and supremum of $g$.$\lim_{a \to 0} x e^{-a x^2} = x$, so we can obtain every value of $\mathbb{R}$. So infimum and supremum are $-\infty$, $+\infty$. Am I right?

Answer 1

Find the derivative of $g$: $$ g'(x)=e^{-ax^2}-2ax^2 e^{-ax^2}=(1-2ax^2)e^{-ax^2}. $$ Since $a$ is positive, the equation $1-2ax^2=0$ has real roots and $x=\pm\sqrt{\frac{1}{2a}}$. Then find these values:

• $\lim_{x\to\infty}g(x)=0$
• $\lim_{x\to-\infty}g(x)=0$
• $g(1/\sqrt{2a})=\frac{1}{\sqrt{2ea}}$: maximum(supremum)
• $g(-1/\sqrt{2a})=-\frac{1}{\sqrt{2ea}}$: minimum(infimum)