which statement is true about the graph of the equation (y = csc^{-1}(x))?\nthere is a horizontal asymptote…

Explanation:

Step1: Recall the domain and range of $y = \csc^{-1}(x)$

The domain of $y=\csc^{-1}(x)$ is $(-\infty,- 1]\cup[1,\infty)$ and the range is $\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]$.

Step2: Analyze horizontal asymptotes

As $x\to\pm\infty$, $\csc^{-1}(x)\to0$. But $y = 0$ is not in the range of $y=\csc^{-1}(x)$. Also, as $x\to\pm\infty$, $y=\csc^{-1}(x)$ does not approach $\frac{\pi}{2}$.

Step3: Analyze vertical asymptotes

The function $y = \csc^{-1}(x)$ is the inverse of $y=\csc(x)$. The function $y = \csc(x)$ has vertical - asymptotes at $x = k\pi,k\in\mathbb{Z}$. For the inverse function $y=\csc^{-1}(x)$, there are no vertical asymptotes at $x = 0$ or $x=\frac{\pi}{2}$. In fact, the function $y=\csc^{-1}(x)$ is continuous on its domain $(-\infty,-1]\cup[1,\infty)$. However, considering the behavior of the inverse - cosecant function, as $x\to\pm\infty$, $y=\csc^{-1}(x)\to0$ but $y = 0$ is not in the range. The correct statement about the asymptotes of $y=\csc^{-1}(x)$ is that there is a horizontal asymptote at $y = 0$.

Answer:

There is a horizontal asymptote at $y = 0$.