91Ó°ÊÓ

When we talk about the domain of a function, we're referring to all possible input values (usually represented as \(x\)) that the function can accept without running into any issues. For a function involving an inverse trigonometric expression like \(f(x)=\sin^{-1}\left(\frac{x}{x+1}\right)\), it is essential to understand the constraints of the trigonometric function itself.
The inverse sine function, \(\sin^{-1}(x)\), only exists for values of \(x\) in the interval \([-1, 1]\). This constraint dictates the domain of the entire function. Therefore, to find the domain of \(f(x)\), we must ensure that \(\frac{x}{x+1}\) remains within this interval.

• First, set up the inequality: \(-1 \leq \frac{x}{x+1} \leq 1\).

This will form the basis for solving which \(x\) values are acceptable inputs for the function.

Inequalities are expressions that involve the symbols \(<\), \(>\), \(\leq\), or \(\geq\) and depict a range of values rather than a specific value. When we solve inequalities, the goal is to find all \(x\) values that satisfy the conditions set by the inequality.

For the given function \(f(x)=\sin^{-1}\left(\frac{x}{x+1}\right)\), we need to solve the following inequalities:

• \(-1 \leq \frac{x}{x+1}\)
• \(\frac{x}{x+1} \leq 1\)

It's important to handle inequalities properly.
This requires understanding how to manipulate and simplify them without altering the meaning. For instance, multiplying or dividing an inequality by a negative number reverses the inequality sign. Inequality solutions tell us the range of \(x\) values that keep the original inequalities true.

Function solving involves a series of steps to find the acceptable input values for a function where it remains defined and within the desired range. After establishing the inequalities \(-1 \leq \frac{x}{x+1}\) and \(\frac{x}{x+1} \leq 1\), we solve each:

• For \(-1 \leq \frac{x}{x+1}\), rearrange and solve for \(x\) by treating \(\frac{x}{x+1}\) as a regular fraction. This might involve cross-multiplying and simplifying.
• For \(\frac{x}{x+1} \leq 1\), perform similar operations to simplify and express \(x\) in isolation.

Next, determine where the solutions overlap, or intersect, which provides the domain of the function.
The intersection signifies the \(x\) values that are valid for both inequalities, ensuring the function is correctly defined over this range.