#6\nidentify the graph of the function ( g(x)=sqrt{-x - 1})\ndetermine when the function is positive…

Explanation:

Step1: Find the domain

For $y = \sqrt{-x - 1}$, the expression under the square - root must be non - negative. So, $-x - 1\geq0$, which gives $x\leq - 1$.

Step2: Determine when the function is positive/negative

Since $y=\sqrt{-x - 1}\geq0$ for all $x$ in its domain, the function is positive when $x\lt - 1$ (because when $x=-1,y = 0$) and non - negative (including zero) for $x\leq - 1$. It is never negative.

Step3: Analyze increasing/decreasing

Let $u=-x - 1$, then $y = \sqrt{u}$. The function $u=-x - 1$ is a linear function with a slope of $-1$, which is decreasing. The function $y=\sqrt{u}$ is increasing for $u\geq0$. By the chain rule, the composition $y=\sqrt{-x - 1}$ is decreasing on its domain $x\leq - 1$.

Step4: Determine end - behavior

As $x\to-\infty$, $u=-x - 1\to\infty$, and $y=\sqrt{-x - 1}\to\infty$.

Answer:

Positive: $x\lt - 1$ Negative: No real numbers Increasing: No real numbers Decreasing: $x\leq - 1$ End behavior: $y\to\infty$ as $x\to-\infty$