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

Decreasing: $(-\infty,0)$ Increasing: nowhere.
$\mathrm{First\:Part:}\:\:$ According to the definitions: A function $\ f\$ defined on an interval is increasing on $\ (a, b)\$ if for every $\ x_1, x_2\$ $\in$ $(a, b)$ $\ x_1\le x_2\$ implies that $\ f(x_1)\le f(x_2).\$ A function $\ f\$ defined on an interval is decreasing on $\ (a, b)\$ if for every $\ x_1, x_2\$ $\in$ $(a, b)$ $\ x_1\le x_2\$ implies that $\ f(x_1)\ge f(x_2).\$ We know that we can take the square root only of positive numbers, i.e $\ \ \sqrt{f(x)}\:\Rightarrow\:f(x)\ge0.$ For the given function $\ y=\sqrt{-x}\$, we have: $-x\ge0$ $\Rightarrow\:x\le0$ $\mathrm{Domain:}\:\:(-\infty,0]$ First of all create a table with a few points to sketch the graph. $\quad \mathrm{See\:the\:table\:and\:graph\:above.}$ $\mathrm{Second\:Part:}\:\:$ The graph does not have any symmetry. $\mathrm{Third\:Part:}\:\:$ The graph of the given function $\ y=\sqrt{-x}\$ is decreasing on $\ (-\infty,0).$