91Ó°ÊÓ

Inverse trigonometric functions are the opposites of the well-known trigonometric functions like sine, cosine, and tangent. They allow us to derive the angle from the ratio of sides in a right triangle. For sine, the function is represented as \( \sin^{-1} \) or \( \arcsin \). This means if \( y = \sin^{-1}(x) \), then \( \sin(y) = x \).
However, unlike sine, whose values repeat over different ranges, the inverse sine function is restricted to a specific domain:

• \( -1 \leq x \leq 1 \) for real inputs.

For example, consider the inverse sine part of our function \( f(x) = \frac{\sin^{-1}(1/x)}{x} \). To have a real number solution from \( \sin^{-1}(1/x) \), the condition

• \( -1 \leq 1/x \leq 1 \)

must be satisfied. This essentially delimits \( x \) to values outside of the interval \(-1 < x < 1\), ensuring the operation is valid.

The domain of a function is the complete set of possible values for the input, in this case, \( x \). Determining the domain involves finding where the function is defined and operates without error.
For \( f(x) = \frac{\sin^{-1}(1/x)}{x} \), let's break it down:

• The expression \( \sin^{-1}(1/x) \) implies \( x \) cannot be zero, as division by zero is undefined.
• Moreover, \( x \) has to satisfy \( x \leq -1 \) or \( x \geq 1 \) due to the restrictions from the inverse sine function.

Combining these, the domain of \( f(x) \) will be the union of sets where both conditions hold true:

• The domain is \( (-\infty, -1] \cup [1, \infty) \).

This union represents all feasible \( x \)-values where the function can be calculated.

Continuity in functions essentially means that small changes in \( x \) do not lead to abrupt changes in \( f(x) \). A function \( f \) is continuous at a point, \( c \), if it is defined at \( c \), the limit as \( x \) approaches \( c \) exists, and that limit equals \( f(c) \).
For \( f(x) = \frac{\sin^{-1}(1/x)}{x} \), we assess continuity over its domain:

• Within \((-\infty, -1)\) and \((1, \infty)\), both \( \sin^{-1}(1/x) \) and \( x \) are continuous functions, meaning their combination also remains continuous.
• At the endpoints \( x = -1 \) and \( x = 1 \), we check for continuity explicitly by examining limits as \( x \) approaches these points from within the domain. Since the limits at \( x = -1 \) and \( x = 1 \) are found to match the function values, \( f(x) \) is indeed continuous at these points too.

Conclusively, there are no sudden jumps or breaks in \( f(x) \) across its evaluated domain \((-\infty, -1] \cup [1, \infty) \).