## University Calculus: Early Transcendentals (3rd Edition)

$$y=\frac{-1}{\sqrt{4-x^2}}$$ 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 and down again but only in the small interval $x\in(-2,2)$, not even touching $-2$ or $2$. - Since $\sqrt{4-x^2}\gt0$ for $x\in(-2,2)$ and $-1\lt0$, the values of $y$ are always negative. The graph is expected to never pass the $Ox$ line into the positive part of the axis. This is true in the graph. - 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 $x\to2$, $y\to-\infty$ again). The graph would go from $-\infty$ near $x=-2$ again to $-\infty$ near $x=2$, but it would not go straight. The graph is narrow, going up from $-\infty$ as $x$ moves away from $-2$, but going down again to $-\infty$ as $x$ approaches $2$. The graph approves of all these predictions.