University Calculus: Early Transcendentals (3rd Edition)

$$y=\frac{x}{\sqrt{4-x^2}}$$ The graph is shown below. From the formula, we takes note of two things: - First, the function is defined as $4-x^2\ge0$, which means $x^2\le4$, so $-2\le x\le 2$. Furthermore, $\sqrt{4-x^2}\ne0$, meaning that $x\ne\pm2$. The domain is limited to $(-2,2)$. We would expect the graph to range only in the small interval $(-2,2)$. This is definitely the case in the graph. The graph takes a curve from below to high above but only in the small interval $x\in(-2,2)$, not even touching $-2$ or $2$. - Second, as $x\to-2$ and $x\to2$, $\sqrt{4-x^2}$ both approaches $0$, meaning the function would approach $\pm\infty$ (in detail, $x\to-2$, $y\to-\infty$, and $x\to2$, $y\to\infty$). The graph would go from $-\infty$ near $x=-2$ to $\infty$ near $x=2$, meaning the graph must be narrow and change quite steeply on the way. This is the case as we look at the graph.