91Ó°ÊÓ

The inverse sine function, often denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), is used to determine the angle whose sine is \( x \). The key characteristic of this function is its domain, which is the set of values \( x \) can have when you plug it into the \( \sin^{-1} \) function. This domain is crucial for solving problems because \( \sin^{-1}(x) \) is only defined for \( x \) values in the interval \([-1, 1]\).

In the given problem, \( f(x)=\sin^{-1}\left(\frac{1-|x|}{2}\right) + \ldots \), we need the internal part, \( \frac{1-|x|}{2} \), to also fit between -1 and 1. To ensure that, we translate the condition \(-1 \leq \frac{1-|x|}{2} \leq 1\) into inequalities that help define the domain of \( x \). This forms the backbone for solving the main problem.

Inequalities are mathematical statements that involve less than (<), greater than (>), or equal to (=) symbols. Solving inequalities involves finding the range of values that satisfy the inequality condition.

For our problem, solving the inequalities derived from \( \sin^{-1}\left(\frac{1-|x|}{2}\right) \) is critical. The expression \(-1 \leq 1 - |x| \leq 1\) is split into two:

• \( -1 \leq 1 - |x| \)
• \( 1 - |x| \leq 1 \)

Depending on whether \(x \) is positive or negative, you solve each part accordingly.

When \(x \geq 0\), \(-1 \leq 1 - x \leq 1\) simplifies to \(0 \leq x \leq 2\), and when \(x < 0\), \(-1 \leq 1 + x \leq 1\) gives \( -2 \leq x \leq 0 \). Combining these solutions we find that \( |x| \) must satisfy \(-2 \leq x \leq 2\). These help us determine the permissible \(x\) values that satisfy the main function's domain constraints.

Rational functions are expressions that involve the division of two polynomials. A crucial property to remember is that the denominator of a rational function cannot be zero, as division by zero is undefined.

In the exercise at hand, the rational function is \( \frac{e^{x}-1}{e^{x}+1} \). The critical part is ensuring the denominator \( e^{x} + 1 \) remains non-zero, so inequality \( e^{x} + 1 eq 0 \) is considered.

Since the exponential function \( e^{x} \) is always positive, \( e^{x} + 1 \) always results in positive numbers greater than or equal to 1 for all real values of \( x \). Hence, this function places no further restriction on \( x \). Thus, the inequality for the numerator is sometimes more relevant to check for additional constraints. However, in this case, it tells us that \( e^{x} \) only affirms our results from prior steps. Together, the rational function and inverse sine constraints conclude \( -2 \leq x \leq 2 \) as the final domain.