Can we sketch the curve $y=e^{-x}/(1+x^2)$?

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We know \(e^x > 0\) for every \(x\), so \(e^{-x}\) has to be always positive too. We also know \(x^2\geq 0\), so \(\dfrac{1}{1+x^2}\) must be always positive.

The product of two positive numbers is also positive, and so \[\dfrac{e^{-x}}{1+x^2}>0,\] for every value of \(x\).

We can find \(y'\) using the quotient rule. We find \[\begin{align*} \dfrac{\mathrm{d}y}{\mathrm{d}x}&= \dfrac{-e^{-x}(1+x^2)-e^{-x}2x}{(1+x^2)^2}\\ &= \dfrac{-e^{-x}(1+x^2+2x)}{(1+x^2)^2}\\ &= \dfrac{-e^{-x}(1+x)^2}{(1+x^2)^2}. \end{align*}\] We have that \(e^{-x}, (1+x)^2\) and \((1+x^2)^2\) are all always positive or zero, and thus \(y'\) is never positive.

We can see that \(y' = 0 \implies x = -1, y = \dfrac{e}{2}.\)

Since \(y' \leq 0\), we know the function is never-increasing, so this point isn’t a maximum or a minimum, and must be a point of inflection.

To sketch the function, it’s useful to calculate a few more characteristics.

We have that the point \((0,1)\) is on the curve; we know from the question that \(y \to \infty\) as \(x \to -\infty\), and we can see that \(y \to 0\) for \(x \to \infty\).