Step-by-Step Solution

Step 1: Understand the function

The function is $f(x) = \sin^{-1}x$, which is the inverse sine function. The domain is restricted to $x \in [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$.

Step 2: Determine key points for graphing

We need to find the values of $f(x)$ at the endpoints of the domain and at some intermediate points.

• When $x = -\frac{1}{\sqrt{2}}$, $f(x) = \sin^{-1}(-\frac{1}{\sqrt{2}}) = -\frac{\pi}{4}$
• When $x = 0$, $f(x) = \sin^{-1}(0) = 0$
• When $x = \frac{1}{\sqrt{2}}$, $f(x) = \sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$

Step 3: Sketch the graph

The graph of $f(x) = \sin^{-1}x$ is a curve that passes through the points $(-\frac{1}{\sqrt{2}}, -\frac{\pi}{4})$, $(0, 0)$, and $(\frac{1}{\sqrt{2}}, \frac{\pi}{4})$. Since the domain is restricted, we only consider the portion of the graph within the given interval.

The graph will be an increasing function within the given domain.

Step 4: Determine the range

The range of $f(x)$ is the set of all possible values of $f(x)$ for $x$ in the given domain. Since $f(x)$ is continuous and increasing on the interval $[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$, the range is the interval between the values of $f(x)$ at the endpoints.

The range is $[-\frac{\pi}{4}, \frac{\pi}{4}]$.