91Ó°ÊÓ

Inverse trigonometric functions play a crucial role in calculus, helping us solve equations involving angles. One of these functions is the inverse sine function, denoted as \( \sin^{-1}(x) \). It helps us find angle values from the ratio of the opposite side to the hypotenuse in a right triangle. In the problem, \( \sin^{-1}(1/x) \) is used, which indicates we need to find an angle whose sine corresponds to \( 1/x \).
Unlike the traditional sine function that works for all real numbers, the inverse sine function has a limited range and domain.

• The domain of \( \sin^{-1}(x) \) is \(-1 \leq x \leq 1\), meaning it can only accept inputs between -1 and 1.
• The range, or all possible output angles, is \( y \) values such that \(-\pi/2 \leq y \leq \pi/2\).

In this exercise, correctly setting the condition \(-1 \leq 1/x \leq 1\) ensures our input for \( \sin^{-1} \) remains within its defined limits. This constraint is crucial for understanding every subsequent step the solution explored. Knowing the domain restrictions of inverse trigonometric functions aids in determining where a function is continuous.

The domain of a function refers to all possible input values for which the function is defined. Determining the domain is a foundational part of understanding functions in calculus. The function given, \( f(x) = \frac{\sin^{-1}(1/x)}{x} \), involves both an inverse trigonometric function and division, each with specific domain considerations.
For \( \frac{\sin^{-1}(1/x)}{x} \) to be defined:

• The expression \( 1/x \) must be within the domain \(-1 \leq 1/x \leq 1\).
• Additionally, \( x \) cannot be zero, because division by zero would make the function undefined.

Solving the inequalities step by step, you identify that \( 1/x \leq 1 \) implies \(-1 \leq x\) and \(-1 \leq 1/x \) suggests \( x \leq -1 \) or \( x \geq 1 \). So, the valid \( x \) values combining these results are \( x \geq 1 \). Therefore, the function is defined for \( x \) belonging to \([1, \infty)\). Understanding a function's domain ensures you're evaluating the function where it's properly defined, crucial for exploring continuity and behavior across different intervals.

Discontinuity occurs in a function where it isn't defined or cannot be computed smoothly. In this exercise, we examined \( f(x) = \frac{\sin^{-1}(1/x)}{x} \) to determine where it might be discontinuous.
For any discontinuity points:

• Consider where the function's components face issues, such as division by zero or exceeding the domain of a function.
• Here, \( f(x) \) is specifically not defined at \( x = 0 \) due to division by zero.

Hence, \( x = 0 \) is a clear point of discontinuity.
Evaluating the rest of the domain, from \( x = 1 \) to infinity, the function remains smooth because each component function—\( \sin^{-1}(1/x) \) and the division by \( x \)—is continuous.
By examining these potential issues, you ensure that you understand how the function behaves and where it may "break," providing insights into the function's overall continuity.