Question

Find DRG of sin^-1sin6/1+x^2

Answer 1

DRG of $f(x) = \sin^{-1}\left(\sin\left(\frac{6}{1+x^2}\right)\right)$:

Domain: $\mathbb{R}$

Range: $[6-2\pi, \frac{\pi}{2}]$

Graph: Symmetric about the y-axis. Defined piecewise: For $|x| \ge \sqrt{\frac{12}{\pi}-1}$, $f(x) = \frac{6}{1+x^2}$. For $\sqrt{\frac{4}{\pi}-1} \le |x| < \sqrt{\frac{12}{\pi}-1}$, $f(x) = \pi - \frac{6}{1+x^2}$. For $0 \le |x| < \sqrt{\frac{4}{\pi}-1}$, $f(x) = \frac{6}{1+x^2} - 2\pi$.

Answer 2

Let $y = \frac{6}{1+x^2}$. The domain of $y$ is $\mathbb{R}$, and the range is $(0, 6]$. The function is $f(x) = \sin^{-1}(\sin y)$. The domain of $f(x)$ is the domain of $y$, which is $\mathbb{R}$. The range of $f(x)$ is the set of values taken by $\sin^{-1}(\sin y)$ for $y \in (0, 6]$. The minimum value of $f(x)$ occurs at $x=0$ where $y=6$, giving $f(0) = \sin^{-1}(\sin 6)$. Since $\frac{3\pi}{2} < 6 < \frac{5\pi}{2}$, $\sin^{-1}(\sin 6) = 6 - 2\pi$. The maximum value of $f(x)$ occurs when $y = \frac{\pi}{2}$ (since $\frac{\pi}{2} \in (0, 6]$ and the maximum value of $\sin^{-1}(\sin y)$ is $\frac{\pi}{2}$), which happens when $\frac{6}{1+x^2} = \frac{\pi}{2} \implies 1+x^2 = \frac{12}{\pi} \implies x^2 = \frac{12}{\pi}-1$. At these values of $x$, $f(x) = \sin^{-1}(\sin \frac{\pi}{2}) = \frac{\pi}{2}$. The range is $[6-2\pi, \frac{\pi}{2}]$. The graph is constructed by considering the piecewise definition of $\sin^{-1}(\sin y)$ based on the intervals of $y$ and mapping them to intervals of $x$. The graph is symmetric about the y-axis.