Study of 1 − e^(−2(x−2)²)

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Study the function $y = 1 - e^{-2(x-2)^2}$:

• study the sign;
• find relative maxima and minima;
• study concavity/convexity and any inflection points.

Solution

Sign and limits. Since $0 < e^{-2(x-2)^2} \le 1$, we have $0 \le y < 1$. Moreover $y = 0$ only at $x = 2$ (where the exponent is zero), while $y > 0$ elsewhere. As $x \to \pm\infty$, $e^{-2(x-2)^2} \to 0$, so $y \to 1$: the line $y = 1$ is a horizontal asymptote ($\sup = 1$, not attained).

Extrema. $y' = 4(x-2)\,e^{-2(x-2)^2}$. The exponential factor is always positive, so $y' = 0 \iff x = 2$; there $y'$ changes from $-$ to $+$: minimum (absolute) at $(2,\,0)$.

Concavity and inflections. $y'' = 4e^{-2(x-2)^2}\bigl(1 - 4(x-2)^2\bigr)$, which vanishes when $(x-2)^2 = \tfrac14$, i.e. $x = \tfrac32$ and $x = \tfrac52$: two inflection points. The function is concave up between the inflections ($\tfrac32 < x < \tfrac52$) and concave down outside. $\boxed{\text{min } (2,\,0);\quad \text{inflections at } x = \tfrac32,\ \tfrac52;\quad \text{asymptote } y = 1}$