91Ó°ÊÓ

When dealing with logarithmic functions, such as \( \ln(1-x) \), understanding when the function is defined is critical. Logarithms require their inside expression, known as the argument, to be positive. In our exercise, we have \( \ln(1-x) \), meaning that the function is only defined when \( 1-x > 0 \). This is because the logarithm of zero or a negative number is not defined in the real number system.

Solving the inequality:

• Start with the condition \( 1-x > 0 \).
• Rearrange it to find \( x < 1 \).

Hence, for the logarithmic part of the function to remain defined, \( x \) must be less than 1. This step ensures that the logarithmic component of our function is safe and sound to use. Focusing on these conditions sets the base for deciding the overall domain of the composite function.

Square root functions, like \( \sqrt{\ln(1-x)} \), have their own domain limitations. They require that the expression under the square root be non-negative because square roots of negative numbers are not real.

Addressing \( \sqrt{\ln(1-x)} \):

• The expression \( \ln(1-x) \) must satisfy \( \ln(1-x) \geq 0 \).

This means that \( 1-x \) must be greater than or equal to 1, as the logarithm of any number less than 1 results in a negative value.

Let’s solve this inequality:

• We get \( \ln(1-x) \geq 0 \), which translates to \( 1-x \geq 1 \).
• This simplifies to \( x \leq 0 \).

Therefore, the square root part limits \( x \) even further, allowing only those values less than or equal to 0 to keep our function defined. It is crucial to consider both the square root and logarithmic restrictions when determining the function's domain.

In problems involving both the logarithm and square root functions, such as \( y = \sqrt{\ln(1-x)} \), solving inequalities is a key skill. Let’s review how solving inequalities helps in finding the domain.

The first inequality comes from the logarithmic function:

• \( 1-x > 0 \)
• This condition gives us \( x < 1 \).

The second inequality arises from ensuring the square root expression remains valid:

• \( \ln(1-x) \geq 0 \)
• This condition refines our range to \( x \leq 0 \).

To find the correct domain, you must satisfy both inequalities simultaneously. This can be visualized as finding the overlap or intersection of allowed \( x \) values. In our solution, this intersection results in \( x \leq 0 \), as it satisfies both of our initial inequalities.

By dissecting and solving these inequalities carefully, you ensure that each part of the function is properly defined, creating a comprehensive understanding of the domain for complex functions.