Problem 56 - Precalculus

Question

Find the domain of the function. \(f(x)=\frac{\sqrt{-1-x}}{\log _{\frac{1}{2}}(x)}\)

Step-by-Step Solution

Answer

The domain of \( f(x) \) is an empty set; there are no values of \( x \) that satisfy both conditions.

Step 1: Understanding the Function

The function is given as \( f(x) = \frac{\sqrt{-1-x}}{\log_{\frac{1}{2}}(x)} \). It combines a square root in the numerator and a logarithm in the denominator. These components have separate constraints that define the domain.

Step 2: Finding Domain for the Square Root

The square root component, \( \sqrt{-1-x} \), is defined only for non-negative arguments. Thus, we need \( -1-x \geq 0 \). Solving this gives \( x \leq -1 \).

Step 3: Finding Domain for the Logarithm

The logarithmic component \( \log_{\frac{1}{2}}(x) \) is defined for positive \( x \). Additionally, since the logarithm is in the denominator, \( \log_{\frac{1}{2}}(x) eq 0 \). To find where the logarithm is zero, set \( \frac{1}{2}^x = 1 \), which happens when \( x = 1 \). Thus, \( x > 0 \) and \( x eq 1 \).

Step 4: Combining Conditions

From the square root condition, \( x \leq -1 \). From the logarithm condition, \( x > 0 \) and \( x eq 1 \). These conditions cannot be true simultaneously. Therefore, there are no values of \( x \) that satisfy both conditions.

Key Concepts

Square Root Function

The square root function is a fundamental concept in mathematics. It involves taking the square root of a number or expression, which essentially finds a value that, when multiplied by itself, gives back the original number. For example, the square root of 9 is 3, because 3 times 3 equals 9. When dealing with square root functions, it's crucial to remember that the expression inside the square root must be non-negative. This is because you cannot take the square root of a negative number in the set of real numbers. For the function in our exercise, represented as \( \sqrt{-1-x} \), the expression \(-1-x\) must be greater than or equal to zero. Solving this inequality, \(-1-x \geq 0\), simplifies to \(x \leq -1\). Therefore, for the square root component of the function to be defined, \(x\) must be less than or equal to \(-1\). This defines the domain for the square root part of the function.

Logarithmic Function

The logarithmic function is the inverse of the exponential function. It helps us solve equations where the unknown appears as the exponent. A logarithm answers the question: "To what power must the base be raised, to produce a given number?"For the exercise, we have the function \( \log_{\frac{1}{2}}(x) \). Here, the base of the logarithm is \(\frac{1}{2}\). Importantly, logarithms are only defined for positive arguments; thus, \(x\) must be greater than zero. Furthermore, because it is in the denominator, \(\log_{\frac{1}{2}}(x)\) should not equal zero.The condition \(\log_{\frac{1}{2}}(x) eq 0\) occurs when \(\frac{1}{2}^x eq 1\). This is true when \(x eq 1\), as \(\frac{1}{2}^1 = \frac{1}{2}\), and it's only when raised to the power of 0 do you get 1 in any base scenario. So, the domain for the logarithmic function requires \(x > 0\) and \(x eq 1\). This limits values to greater than zero except for one instead of all real numbers.

Inequalities

Inequalities are a central part of solving domain problems in functions. They represent a range of values that are possible for a variable, rather than a single value. This is because many functions, including our square root and logarithmic functions, have restrictions on the values they can take.The inequality for the square root \(-1-x \geq 0\) indicates that \(-1-x\) can be zero or any positive number. Solving such inequality yields \(x \leq -1\). This means that \(x\) must satisfy this condition for the square root term to be defined.However, for the logarithmic part, two inequalities are relevant: \(x > 0\) and \(x eq 1\). These two restrict \(x\) to be in specific areas of the real number line. Combining these inequalities means looking for values of \(x\) that satisfy all conditions simultaneously, which in this exercise is unfeasible, rendering the function's overall domain as empty.

Precalculus Concepts

Precalculus forms the foundation for understanding more advanced mathematical subjects like calculus, and it includes concepts from algebra and trigonometry. This is crucial for tackling problems involving function domains as it combines these fundamental ideas.In evaluating the domain of a function like \( f(x) = \frac{\sqrt{-1-x}}{\log_{\frac{1}{2}}(x)} \), we use knowledge of different types of functions and how their domains are restricted. The problem involves:

• Understanding function restrictions such as square roots requiring non-negative numbers.
• Logarithms only accepting positive arguments and avoiding zero as a denominator.
• Solving inequalities, which is crucial for defining where these restrictions apply to \(x\).
• Combining all the obtained conditions to understand their intersection.

These precalculus concepts must be mastered to grasp why the domain for such a composite function might become nonexistent, as the conditions cannot hold true at the same time.