Study of x·e^(−x²/2)

Text

Study the function $y = x\,e^{-x^2/(2a)}$ (with $a > 0$ a parameter):

• (a) compute the derivative (with $a$ as a parameter), then set $a = 1$;
• (b) study the sign;
• (c) find relative maxima and minima, discussing also absolute maxima/minima and supremum/infimum;
• (d) sketch the graph.

Solution

(a) Derivative. In general $y' = e^{-x^2/(2a)}\left(1 - \frac{x^2}{a}\right).$ With $a = 1$: $\ y = x\,e^{-x^2/2}$ and $\ y' = e^{-x^2/2}\,(1 - x^2)$.

(b) Sign. Since $e^{-x^2/2} > 0$, the sign of $y$ equals the sign of $x$: $\ y > 0$ for $x > 0$, $\ y < 0$ for $x < 0$. The function is odd ($y(-x) = -y(x)$).

(c) Extrema. $y' = 0 \iff 1 - x^2 = 0 \iff x = \pm 1$.

• At $x = 1$: maximum, $f(1) = e^{-1/2} = \tfrac{1}{\sqrt e} \approx 0{,}6065$.
• At $x = -1$: minimum, $f(-1) = -\tfrac{1}{\sqrt e} \approx -0{,}6065$. As $x \to \pm\infty$, $y \to 0$: the line $y = 0$ is a horizontal asymptote. Given the behaviour, the relative maximum is also the absolute maximum and the relative minimum the absolute minimum; hence $\sup = \tfrac{1}{\sqrt e}$ (attained) and $\inf = -\tfrac{1}{\sqrt e}$ (attained).

(d) Concavity. $y'' = x(x^2 - 3)\,e^{-x^2/2}$, which vanishes at $x = 0$ and $x = \pm\sqrt 3 \approx \pm 1{,}732$: three inflection points. $\boxed{\text{max } \left(1,\ \tfrac{1}{\sqrt e}\right);\quad \text{min } \left(-1,\ -\tfrac{1}{\sqrt e}\right);\quad \text{inflections at } x = 0,\ \pm\sqrt 3}$