Review question

# Can we sketch the curve $y=e^{-x}/(1+x^2)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9484

## Solution

If $y=\dfrac{e^{-x}}{1+x^2}$, prove that (i) $y$ is always positive, …

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 $x$.

…, (ii) $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is never positive.

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.

Find the coordinates of the point where $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$, and sketch the graph of the function.

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 $x \to \infty$.